Interview question: data structure to set all values in O(1)
I encountered the following interview question on the Internet.
Describe a data structure for which getValue(int index), setValue(int index, int value), and setAllValues(int value) are all O(1).
Though array is good enough for the first and second operations to be performed in O(1), what can be proposed for the third (setAllValues)?
data-structures
add a comment |
I encountered the following interview question on the Internet.
Describe a data structure for which getValue(int index), setValue(int index, int value), and setAllValues(int value) are all O(1).
Though array is good enough for the first and second operations to be performed in O(1), what can be proposed for the third (setAllValues)?
data-structures
1
@leppie I disagree - see my answer :)
– Timothy Jones
Apr 4 '12 at 6:05
1
@TimothyJones: OK, I understand your answer, but it is 'cheating'. You are only setting 1 value, not all values.
– leppie
Apr 4 '12 at 6:08
3
@leppie The data structure would work as specified, so I'm not sure what's cheating about it ;)
– Timothy Jones
Apr 4 '12 at 7:27
Is it feasible to implement by a hash table?
– Hengameh
May 2 '15 at 8:15
@RoyiNamir: But as soon as you set a value again, you get the old values back in your code.
– leppie
Jan 10 '16 at 19:05
add a comment |
I encountered the following interview question on the Internet.
Describe a data structure for which getValue(int index), setValue(int index, int value), and setAllValues(int value) are all O(1).
Though array is good enough for the first and second operations to be performed in O(1), what can be proposed for the third (setAllValues)?
data-structures
I encountered the following interview question on the Internet.
Describe a data structure for which getValue(int index), setValue(int index, int value), and setAllValues(int value) are all O(1).
Though array is good enough for the first and second operations to be performed in O(1), what can be proposed for the third (setAllValues)?
data-structures
data-structures
edited Nov 11 '18 at 3:33
Cœur
17.7k9106145
17.7k9106145
asked Apr 4 '12 at 5:44
AlecAlec
7262927
7262927
1
@leppie I disagree - see my answer :)
– Timothy Jones
Apr 4 '12 at 6:05
1
@TimothyJones: OK, I understand your answer, but it is 'cheating'. You are only setting 1 value, not all values.
– leppie
Apr 4 '12 at 6:08
3
@leppie The data structure would work as specified, so I'm not sure what's cheating about it ;)
– Timothy Jones
Apr 4 '12 at 7:27
Is it feasible to implement by a hash table?
– Hengameh
May 2 '15 at 8:15
@RoyiNamir: But as soon as you set a value again, you get the old values back in your code.
– leppie
Jan 10 '16 at 19:05
add a comment |
1
@leppie I disagree - see my answer :)
– Timothy Jones
Apr 4 '12 at 6:05
1
@TimothyJones: OK, I understand your answer, but it is 'cheating'. You are only setting 1 value, not all values.
– leppie
Apr 4 '12 at 6:08
3
@leppie The data structure would work as specified, so I'm not sure what's cheating about it ;)
– Timothy Jones
Apr 4 '12 at 7:27
Is it feasible to implement by a hash table?
– Hengameh
May 2 '15 at 8:15
@RoyiNamir: But as soon as you set a value again, you get the old values back in your code.
– leppie
Jan 10 '16 at 19:05
1
1
@leppie I disagree - see my answer :)
– Timothy Jones
Apr 4 '12 at 6:05
@leppie I disagree - see my answer :)
– Timothy Jones
Apr 4 '12 at 6:05
1
1
@TimothyJones: OK, I understand your answer, but it is 'cheating'. You are only setting 1 value, not all values.
– leppie
Apr 4 '12 at 6:08
@TimothyJones: OK, I understand your answer, but it is 'cheating'. You are only setting 1 value, not all values.
– leppie
Apr 4 '12 at 6:08
3
3
@leppie The data structure would work as specified, so I'm not sure what's cheating about it ;)
– Timothy Jones
Apr 4 '12 at 7:27
@leppie The data structure would work as specified, so I'm not sure what's cheating about it ;)
– Timothy Jones
Apr 4 '12 at 7:27
Is it feasible to implement by a hash table?
– Hengameh
May 2 '15 at 8:15
Is it feasible to implement by a hash table?
– Hengameh
May 2 '15 at 8:15
@RoyiNamir: But as soon as you set a value again, you get the old values back in your code.
– leppie
Jan 10 '16 at 19:05
@RoyiNamir: But as soon as you set a value again, you get the old values back in your code.
– leppie
Jan 10 '16 at 19:05
add a comment |
11 Answers
11
active
oldest
votes
How about an array
of tuples timestamp, value
, with an additional timestamp, value
called all
. Since you only care about the relative time of insertions, you can use a monotonically increasing id
for the values of timestamp:
type Entry
int timestamp,
int value
type structure
int id
Entry all
Entry array
Initialise all fields to 0. Then the following should work for you:
setValue(index i, value v):
array[i] = id++, value
value getValue(index i)
if(all.timestamp > array[i].timestamp) return all.value
else return array[i].valuesetAll(value v)
all = id++, value
A problem with this approach is that eventually you'll run out of ids for timestamp, and might wrap around. If you chose a 64 bit value to store timestamps, then this gives you 18,446,744,073,709,551,616 insertions or setAlls before this happens. Depending on the expected use of the datastructure, an O(n) cleanup phase might be appropriate, or you could just throw an exception.
Another issue that might need to be considered is multi-threading. Three obvious problems:
- if
id++
isn't atomic and two threads obtained a newid
at the same time then you could get two entries with the same id. - if the incrementation of id and the assignment of the created tuple aren't atomic together (they're probably not) and there were two simultaneous inserts, then you could get a race condition where the older id wins.
- if the assignment of the tuple isn't atomic, and there's an
insert()
at the same time as aget()
then you might end up in a position where you've got saynew_id, old_value
in the array, causing the wrong value to be returned.
If any of these are problems, the absolute easiest solution to this is to put "not thread safe" in the documentation (cheating). Alternatively, if you can't implement the methods atomically in your language of choice, you'd need to put some sort of synchronisation locks around them.
But we can;t be sure that how it works internally. This single statement can be O(n).
– ganesshkumar
Apr 4 '12 at 6:05
1
This relies on every call tocurrent_time()
returning a different value though (otherwise, two timestamps may be equal)
– Damien_The_Unbeliever
Apr 4 '12 at 6:07
@Damien_The_Unbeliever that's true. You may also have to deal with wrap-around cases too, if the program was around for a while.
– Timothy Jones
Apr 4 '12 at 6:15
Thanks Timothy. Perfect!
– Alec
Apr 4 '12 at 6:42
1
Your signature forgetValue()
needs to be fixed.
– Patrick Roberts
Dec 18 '17 at 7:57
|
show 6 more comments
I got the same question in one of the technical interviews.
Here is my complete ready-to-use Java-implementation including the test cases.
The key idea is to keep the value of setAll()
in special variable (e.g. joker
) and then trace the change of this value in a proper way.
In order to keep the code clean, some access modifiers have been abolished.
Node class:
import java.time.LocalDate;
class Node
int value;
LocalDate jokerSetTime;
Node(int val, LocalDate jokSetTime)
value = val;
jokerSetTime = jokSetTime;
DS class:
class DS
Node arr;
DS(int len)
arr = new Node[len];
DataStructure class:
import java.time.LocalDate;
class DataStructure
private boolean isJokerChanged;
private Integer joker;
private LocalDate jokerSetTime;
private DS myDS;
DataStructure(int len)
myDS = new DS(len);
Integer get(int i)
Integer result;
if (myDS.arr.length < i)
return null;
// setAll() has been just called
if (isJokerChanged)
return joker;
if (myDS.arr[i] == null)
// setAll() has been previously called
if (joker != null)
result = joker;
else
result = null;
else if (myDS.arr[i].jokerSetTime == jokerSetTime)
// cell value has been overridden after setAll() call
result = myDS.arr[i].value;
else
result = joker;
return result;
void setAll(int val)
isJokerChanged = true;
joker = val;
jokerSetTime = LocalDate.now();
void set(int i, int val)
isJokerChanged = false;
myDS.arr[i] = new Node(val, jokerSetTime);
Main class:
class Main
public static void main(String args)
DataStructure ds = new DataStructure(100);
Integer res;
res = ds.get(3);
ds.set(3, 10);
res = ds.get(3);
ds.setAll(6);
res = ds.get(3);
res = ds.get(15);
ds.set(4, 7);
res = ds.get(4);
res = ds.get(3);
ds.setAll(6);
ds.set(8, 2);
res = ds.get(3);
Update:
The code has been updated. The previous implementation didn't take into account the case when setAll()
is called twice in a row with the same value and is followed by set(x)
, get(y)
, e.g.: setAll(100)
, set(3, 1)
, setAll(100)
, set(5, 3)
, get(3)
.
The use of timestamp approach has been added to allow distinguishing between different setAll()
calls with identical values.
P.S. This implementation is not a thread safe.
2
I have a strong hunch that this was expected. All the timestamp based solutions add completely unnecessary complexity.
– user4815162342
Apr 10 '18 at 14:15
2
@MikeB. Great implementation! Very clean and clear. At first I was having trouble understanding the previous solutions with all of the timestamps, but this is just amazing, exactly what I was looking for. Thanks again!
– Martin
Jun 10 '18 at 9:54
1
This solution is wrong. Consider the following example in the main function:ds.setAll(6); ds.set(3,10); ds.setAll(6); ds.set(4,1); res = ds.get();
It will return the incorrect value of 10, even thoughsetAll(6)
was called after it. The secondds.set()
call is needed to clear theisJokerChanged
variable, which would have superficially defended against this bug otherwise.
– AnatolyVorobey
Jul 12 '18 at 17:29
@AnatolyVorobey, the code has been fixed. Thanks for the test case.
– Mike B.
Jul 15 '18 at 12:34
add a comment |
How about an array of pointers to a single common value? Set the value, and all references will point to the single changed value in O(1)..
Answers the question though! ;) lol.
– WOPR
Apr 4 '12 at 6:07
This works until you want to change one of the elements to point to something other than the common value.
– Timothy Jones
Apr 4 '12 at 6:09
That wasn't a requirement specified in the question.
– WOPR
Apr 4 '12 at 6:10
1
I think it is - if you call setAll(), and then setValue(), you'd have to change one of the elements.
– Timothy Jones
Apr 4 '12 at 6:16
3
I would say that specs clearly imply that it should be possible to set values individually without affecting the other values. Otherwise the index parameter to the setValue method is superfluous. And if you do not want to interpret method signatures as specifications, then it would be possible to simply implement all of the methods as NoOps...
– Alderath
Apr 4 '12 at 8:57
add a comment |
I was just asked this question very recently in an Interview. I came up with a hash table implementation. It would solve the problem of running out of the timestamp values but the thread safety feature needs to be implemented (probably using lazy initialization techniques)
Let's say in our class we have a private variable _defaultValue to hold a default value and we also have a private hashtable or dictionary _hashtable.
SetAllValues could just set _defaultValue equal to the value passed and _hashtable initialized/set to a new hashtable and discard any reference to the old hash table. SetValue should just add a new value to _hashtable or update the value if the key ( or index) is already present in the _hashtable. GetValue should check whether the key (or index) is present in _hashtable, then return it, else return the value stored in _defaultValue.
This is my first answer on StackOverflow. I am a little lazy in writing up the code. Will probably edit the answer soon.
The interviewer nodded yes to this solution but insisted to implement it without using a hashtable. I guess, he was asking me to give a similar approach as Timothy's answer. And I wasn't able to get it at that moment :(. Anyways, Cheers!
EDIT:
Posting the code (in C#)
class MyDataStruct
private int _defaultValue;
private Dictionary<int,int> _hashtable;
public MyDataStruct()
_defaultValue = 0; // initialize default with some value
_hashtable = new Dictionary<int, int>();
public void SetAllValues(int val)
_defaultValue = val;
_hashtable = new Dictionary<int, int>();
public void SetValue(int index, int val)
if (_hashtable.ContainsKey(index))
_hashtable.Add(index, val);
else
_hashtable[index] = val;
public int GetValue(int index)
if (_hashtable.ContainsKey(index))
return _hashtable[index];
else
return _defaultValue;
could you please provide your code with hash table?
– Hengameh
May 2 '15 at 6:07
1
@Hengameh -- added!
– tapas nayak
Jul 20 '15 at 6:30
1
This is almost correct except that GetValue for a non existing hey would always return default value even when this key was not inserted to the data structure.
– akrabi
Nov 18 '15 at 18:11
add a comment |
We can have a variable V which stores an int and an array of containing Tuple as Value, id..
And a global int variable G (which will act like identity in SQL and whenever any set or setAll operation is done its value get increment by 1)
initial all Ids and V value will be default say null..
so V = null All Tuple = null, null
set(index i, val v) -> G = G+1, arr[i].Val = v, arr[i].Id = G
get(index i) -> if V.Id > arr[i].Id return V.Val else return arr[i]
set-all(val v) -> G= G+1, V.Id= G, V.Val = v
add a comment |
All existing answers use a timestamp that is incremented on every setVal
operation. This is not necessary. In fact, it is necessary only to increment the timestamp on setAll
. Another issue some raised was arithmetic overflow. This can be handled without breaking the constant time bounds by updating a single cell on each setAll
and performing the time comparison carefully.
How it works
The basic concept is essentially similar to that of the other answers, but with a twist.
What they do: Store the value used for the last setAll
operation separately, and keep track of the time that operation was performed. Each time they perform a setVal
, they store the current time along with the given value in the array. Each time they perform a getVal
, they compare the time in the given location to the time the last setAll
occurred, and then choose either the value at the location or the setAll
value depending on which is greater.
Why this can be a problem: Suppose the current time overflows and a setAll
operation occurs soon afterward. It will appear as though most of the stored array values are newer than the setAll
value, when they are actually older.
The solution: Stop imagining that we're keeping track of the total amount of time that has passed since data structure initialization. Imagine a giant clock with a "second hand" that ticks not 60 times around the circle but rather 2^n times around the circle. The position of the second hand represents the time of the most recent setAll
operation. Each setVal
operation stores this time along with a value. So if we perform a setAll
when the "clock" is at 45, and then perform six setVal
operations on different elements, the setAll
time and the times at all six locations will be the same. We wish to maintain the following invariant:
The time at a given element location equals the
setAll
time if and only if that element was set withsetVal
more recently than the lastsetAll
operation.
Clearly, the procedure described above automatically ensures that if the element was set recently, then its time will equal the setAll
time. The challenge is to make the reverse implication hold as well.
To be continued ....
The code
I've written this in Haskell because that is the language I know best, but it is not exactly the most natural language for the job.
-# LANGUAGE BangPatterns #-
module RepArr where
import Control.Monad.Primitive
import Data.Primitive.MutVar
import qualified Data.Vector.Mutable as V
import Data.Vector.Mutable (MVector)
import Control.Applicative
import Prelude hiding (replicate)
import Control.Monad.ST
import Data.Word
-- The Int in the MutVar is the refresh pointer
data RepArr s a = RepArr (MutVar s (Word, a, Int)) (MVector s (Word,a))
-- Create a fancy array of a given length, initially filled with a given value
replicate :: (PrimMonad m, Applicative m) => Int -> a -> m (RepArr (PrimState m) a)
replicate n a = RepArr <$> newMutVar (0,a,0) <*> V.replicate n (0,a)
getVal :: PrimMonad m => RepArr (PrimState m) a -> Int -> m a
getVal (RepArr allv vec) n = do
(vectime, vecval) <- V.read vec n
(alltime, allval, _) <- readMutVar allv
if (fromIntegral (alltime - vectime) :: Int) > 0
then return allval
else return vecval
setVal :: PrimMonad m => RepArr (PrimState m) a -> Int -> a -> m ()
setVal (RepArr allv vec) i v = do
(!alltime, _, _) <- readMutVar allv
V.write vec i (alltime, v)
setAll :: PrimMonad m => RepArr (PrimState m) a -> a -> m ()
setAll r@(RepArr allv vec) v = do
(oldt, _, oldc) <- readMutVar allv
getVal r oldc >>= setVal r oldc
let !newc = case oldc+1 of
op1 | op1 == V.length vec -> 0
| otherwise -> op1
let !newt = oldt+1
writeMutVar allv (newt, v, newc)
To avoid potential (rare) garbage collection pauses, it's actually necessary to unbox the Int
and Word
values, as well as using unboxed vectors instead of polymorphic ones, but I'm not really in the mood and this is a fairly mechanical task.
Here's a version in C (completely untested):
#include <malloc.h>
struct Pair
unsigned long time;
void* val;
;
struct RepArr
unsigned long allT;
void* allV;
long count;
long length;
struct Pair vec;
;
struct RepArr *replicate (long n, void* val)
struct RepArr *q = malloc (sizeof (struct RepArr)+n*sizeof (struct Pair));
q->allT = 1;
q->allV = val;
q->count = 0;
q->length = n;
int i;
for (i=0; i<n; i++)
struct Pair foo = 0,val;
q->vec[i] = foo;
return q;
void* getVal (struct RepArr *q, long i)
if ((long)(q->vec[i].time - q->allT) < 0)
return q->allV;
else
return q->vec[i].val;
void setVal (struct RepArr *q, long i, void* val)
q->vec[i].time = q->allT;
q->vec[i].val = val;
void setAll (struct RepArr *q, void* val)
setVal (q, q->count, getVal (q, q->count));
q->allV = val;
q->allT++;
q->count++;
if (q->count == q->length)
q->count = 0;
add a comment |
/*
At the time I am writing this, all of the solutions on this page will double (or
more) the amount of space required to store the array. The following solution
reduces the amount of wasted space from Ω(n) to θ(n/w), where w is the number of
bits in a computer "word". On my machine, that's 64.
This prose in this answer is in C comments, so you can copy-and-paste this
answer verbatim and compile it with your C compiler.
*/
#include <stdbool.h>
#include <stddef.h>
#include <stdint.h>
#include <stdlib.h>
/*
The problem is to support reading and writing values in an array in O(1) time
along with bulk writes in which all values in the array are written at once in
O(1) time. This is possible using a technique invented by Aho, Hopcroft and
Ullman, as far as I know. I will present a version due to Gonzalo Navarro,
"Constant-Time Array Initialization in Little
Space".
The idea is to keep three metadata arrays along with the data array. We also
keep two integers: unset
, which is the last value used in a bulk write
operation, and size
, an approximation for the number of values that have been
set since the last bulk write. At all times, the number of distinct values
written since the last bulk write is between size
and w * size
.
The three metadata arrays describe information about blocks of w values in the
data array. They are:
nth
: nth[i] is the ith unique block to be written to since the last bulk
writeinverse_nth
: inverse_nth[i] is the order in in which the ith block of the
array was written, counting from 0 at the last bulk write.bitset
: The jth bit ofbitset[i]
is 1 when the array cell numbered
64*i + j has been written to since the last bulk write.
bitset[i]
and inverse_nth[i]
are allowed to be invalid if i
is not a
member of the set nth[0]
, nth[1]
, ... , nth[size-1]
. In other words,inverse_nth[i]
and bitset[i]
are valid if and only if inverse_nth[i] < size
and nth[inverse_nth[i]] == i
.
Rather than store three separate arrays of the same length, I chose to store one
array, is_set
, with three fields.
*/
typedef struct
int nth_;
int inverse_nth_;
uint64_t bitset_;
IsSetCell;
typedef struct
int unset_;
int size_;
IsSetCell is_set_;
IsSetArray;
typedef struct
IsSetArray * metadata_;
int payload_;
ResettableArray;
/*
To construct an array, we need a default value to return when the reading a
value that has never been written to.
*/
ResettableArray * ConstructResettableArray(int n, int unset)
ResettableArray* result =
malloc(offsetof(ResettableArray, payload_) + n * sizeof(int));
if (!result) return NULL;
n = (n + 63) / 64;
result->metadata_ =
malloc(offsetof(IsSetArray, is_set_) + n * sizeof(IsSetCell));
if (!result->metadata_)
free(result);
return NULL;
result->metadata_->size_ = 0;
result->metadata_->unset_ = unset;
return result;
void DestructResettableArray(ResettableArray * a)
if (a) free(a->metadata_);
free(a);
/*
The bulk of the algorithm is in writing and reading the metadata. AfterIsSet()
and Set()
are defined (below), reading and writing the arrays is
straightforward.
*/
bool IsSet(const IsSetArray * a, int i);
void Set(IsSetArray * a, int i);
int GetValue(const ResettableArray * a, int i)
if (!IsSet(a->metadata_, i)) return a->metadata_->unset_;
return a->payload_[i];
void SetValue(ResettableArray * a, int i, int v)
a->payload_[i] = v;
Set(a->metadata_, i);
void SetAllValues(ResettableArray * a, int v)
a->metadata_->unset_ = v;
/*
The complex part of reading and writing is the bidirectional relationship
between inverse_nth
and nth
. If they point to each other at location i
(is_set[is_set[i].inverse_nth].nth == i
) then location i contains valid data
that was written after the last bulk write, as long as is_set[i].inverse_nth <
.
size
*/
uint64_t OneBit(int i)
return UINT64_C(1) << i;
bool IsSet(const IsSetArray * a, int i)
const int cell = i/64, offset = i & 63;
const int inverse_nth = a->is_set_[cell].inverse_nth_;
return inverse_nth < a->size_ && a->is_set_[inverse_nth].nth_ == cell &&
a->is_set_[cell].bitset_ & OneBit(offset);
void Set(IsSetArray * a, int i) a->is_set_[inverse_nth].nth_ != cell)
a->is_set_[cell].inverse_nth_ = a->size_;
a->is_set_[cell].bitset_ = 0;
a->is_set_[a->size_].nth_ = cell;
++a->size_;
a->is_set_[cell].bitset_
The code never decreases size_, which I think must be wrong. Perhaps it should be zeroed in SetAllValues()?
– AnatolyVorobey
Jul 13 '18 at 6:08
add a comment |
Python example
class d:
def __init__(self, l):
self.len = l
self.g_p = [i for i in range(self.len)]
self.g_v = [0 for i in range(self.len)]
self.g_s = self.len - 1
self.g_c = 0
def getVal(self, i):
if (i < 0 or i >= self.len):
return
if (self.g_p[i] <= self.g_s):
return self.g_v[self.g_p[i]]
return self.g_c
def setVal(self, i, v):
if (i < 0 or i >= self.len):
return
if (self.g_p[i] > self.g_s):
self.g_s += 1
self.g_p[self.g_s], self.g_p[i] = self.g_p[i], self.g_p[self.g_s]
self.g_v[self.g_p[i]] = v
def setAll(self, v):
self.g_c = v
self.g_s = -1
add a comment |
Regarding Timothy Jone's answer:
A problem with this approach is that eventually you'll run out of ids for timestamp, and might wrap around. If you chose a 64 bit value to store timestamps, then this gives you 18,446,744,073,709,551,616 insertions or setAlls before this happens. Depending on the expected use of the datastructure, an O(n) cleanup phase might be appropriate, or you could just throw an exception.
This is exactly the worst case scenario which make this solution O(n) too, and not O(1). This stucture, although saves a lot of potential O(n) insert operation, is still in O(n) effeciency.
See my answer for a way to maintain constant time updates in the face of overflow.
– dfeuer
Mar 6 '15 at 1:51
@dfeuer can you please specify in words how you unbox the Int and Word values? This question is being asked around the world in interviews, and the expected answer by interviewers (as descibed by Timothy) is definitly incorrect, or at least incomplete. In any case, it is obvious that the answer above should be refined. Thanks in advance.
– Emliti
Mar 6 '15 at 13:31
That sounds like a pretty good reason to delete my answer altogether.
– dfeuer
Mar 6 '15 at 13:59
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post.
– Ian
Mar 6 '15 at 14:20
@Bluehound Because I ususally don't write here I cannot leave a comment under Timothy's answer. I know I have made an important remark so please don't delete it. If pereferable move this answer under Timothy's answer so anyone can see and comment. Thank you
– Emliti
Mar 6 '15 at 19:20
|
show 22 more comments
This is my answer in Java (I'm not completly sure about the syntax).
I did the set-functions synchronized in order to avoid a situation where changeT and defChangeT are equal.
Struct ValueSt
int value;
Timestamp changeT;
Class MyDS
int default;
Timestamp defChangeT;
HashMap map;
public MyDS()
map = new HashMap<int, ValueSt>();
public synchronized void mySet(Int key, int value)
ValueSt vs = new ValueSt(value, Timestamp(System.current));
map.put(key, vs);
public synchronized void setAll(int value)
default = value;
defChangeT = Timestamp(System.current));
public int myGet(int key)
ValueSt vs = map.get(key);
if(vs != null)
if(vs.changeT > defChangeT)
return vs.value;
else
return default;
return null;
add a comment |
The correct solution in C#:
public sealed class SpecialDictionary<T, V>
private Dictionary<T, Tuple<DateTime, V>> innerData;
private Tuple<DateTime, V> setAllValue;
private DateTime prevTime;
public SpecialDictionary()
innerData = new Dictionary<T, Tuple<DateTime, V>>();
public void Set(T key, V value) => innerData[key] = new Tuple<DateTime, V>(GetTime(), value);
public void SetAll(V value) => setAllValue = new Tuple<DateTime, V>(GetTime(), value);
public V Get(T key)
Tuple<DateTime, V> tmpValue = innerData[key];
if (setAllValue?.Item1 > tmpValue.Item1)
return setAllValue.Item2;
else
return tmpValue.Item2;
private DateTime GetTime()
if (prevTime == null)
prevTime = DateTime.Now;
else
if (prevTime == DateTime.Now)
Thread.Sleep(1);
prevTime = DateTime.Now;
return prevTime;
And the test:
static void Main(string args)
SpecialDictionary<string, int> dict = new SpecialDictionary<string, int>();
dict.Set("testA", 1);
dict.Set("testB", 2);
dict.Set("testC", 3);
Console.WriteLine(dict.Get("testC"));
dict.SetAll(4);
dict.Set("testE", 5);
Console.WriteLine(dict.Get("testC"));
Console.WriteLine(dict.Get("testE"));
Console.ReadKey();
add a comment |
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11 Answers
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How about an array
of tuples timestamp, value
, with an additional timestamp, value
called all
. Since you only care about the relative time of insertions, you can use a monotonically increasing id
for the values of timestamp:
type Entry
int timestamp,
int value
type structure
int id
Entry all
Entry array
Initialise all fields to 0. Then the following should work for you:
setValue(index i, value v):
array[i] = id++, value
value getValue(index i)
if(all.timestamp > array[i].timestamp) return all.value
else return array[i].valuesetAll(value v)
all = id++, value
A problem with this approach is that eventually you'll run out of ids for timestamp, and might wrap around. If you chose a 64 bit value to store timestamps, then this gives you 18,446,744,073,709,551,616 insertions or setAlls before this happens. Depending on the expected use of the datastructure, an O(n) cleanup phase might be appropriate, or you could just throw an exception.
Another issue that might need to be considered is multi-threading. Three obvious problems:
- if
id++
isn't atomic and two threads obtained a newid
at the same time then you could get two entries with the same id. - if the incrementation of id and the assignment of the created tuple aren't atomic together (they're probably not) and there were two simultaneous inserts, then you could get a race condition where the older id wins.
- if the assignment of the tuple isn't atomic, and there's an
insert()
at the same time as aget()
then you might end up in a position where you've got saynew_id, old_value
in the array, causing the wrong value to be returned.
If any of these are problems, the absolute easiest solution to this is to put "not thread safe" in the documentation (cheating). Alternatively, if you can't implement the methods atomically in your language of choice, you'd need to put some sort of synchronisation locks around them.
But we can;t be sure that how it works internally. This single statement can be O(n).
– ganesshkumar
Apr 4 '12 at 6:05
1
This relies on every call tocurrent_time()
returning a different value though (otherwise, two timestamps may be equal)
– Damien_The_Unbeliever
Apr 4 '12 at 6:07
@Damien_The_Unbeliever that's true. You may also have to deal with wrap-around cases too, if the program was around for a while.
– Timothy Jones
Apr 4 '12 at 6:15
Thanks Timothy. Perfect!
– Alec
Apr 4 '12 at 6:42
1
Your signature forgetValue()
needs to be fixed.
– Patrick Roberts
Dec 18 '17 at 7:57
|
show 6 more comments
How about an array
of tuples timestamp, value
, with an additional timestamp, value
called all
. Since you only care about the relative time of insertions, you can use a monotonically increasing id
for the values of timestamp:
type Entry
int timestamp,
int value
type structure
int id
Entry all
Entry array
Initialise all fields to 0. Then the following should work for you:
setValue(index i, value v):
array[i] = id++, value
value getValue(index i)
if(all.timestamp > array[i].timestamp) return all.value
else return array[i].valuesetAll(value v)
all = id++, value
A problem with this approach is that eventually you'll run out of ids for timestamp, and might wrap around. If you chose a 64 bit value to store timestamps, then this gives you 18,446,744,073,709,551,616 insertions or setAlls before this happens. Depending on the expected use of the datastructure, an O(n) cleanup phase might be appropriate, or you could just throw an exception.
Another issue that might need to be considered is multi-threading. Three obvious problems:
- if
id++
isn't atomic and two threads obtained a newid
at the same time then you could get two entries with the same id. - if the incrementation of id and the assignment of the created tuple aren't atomic together (they're probably not) and there were two simultaneous inserts, then you could get a race condition where the older id wins.
- if the assignment of the tuple isn't atomic, and there's an
insert()
at the same time as aget()
then you might end up in a position where you've got saynew_id, old_value
in the array, causing the wrong value to be returned.
If any of these are problems, the absolute easiest solution to this is to put "not thread safe" in the documentation (cheating). Alternatively, if you can't implement the methods atomically in your language of choice, you'd need to put some sort of synchronisation locks around them.
But we can;t be sure that how it works internally. This single statement can be O(n).
– ganesshkumar
Apr 4 '12 at 6:05
1
This relies on every call tocurrent_time()
returning a different value though (otherwise, two timestamps may be equal)
– Damien_The_Unbeliever
Apr 4 '12 at 6:07
@Damien_The_Unbeliever that's true. You may also have to deal with wrap-around cases too, if the program was around for a while.
– Timothy Jones
Apr 4 '12 at 6:15
Thanks Timothy. Perfect!
– Alec
Apr 4 '12 at 6:42
1
Your signature forgetValue()
needs to be fixed.
– Patrick Roberts
Dec 18 '17 at 7:57
|
show 6 more comments
How about an array
of tuples timestamp, value
, with an additional timestamp, value
called all
. Since you only care about the relative time of insertions, you can use a monotonically increasing id
for the values of timestamp:
type Entry
int timestamp,
int value
type structure
int id
Entry all
Entry array
Initialise all fields to 0. Then the following should work for you:
setValue(index i, value v):
array[i] = id++, value
value getValue(index i)
if(all.timestamp > array[i].timestamp) return all.value
else return array[i].valuesetAll(value v)
all = id++, value
A problem with this approach is that eventually you'll run out of ids for timestamp, and might wrap around. If you chose a 64 bit value to store timestamps, then this gives you 18,446,744,073,709,551,616 insertions or setAlls before this happens. Depending on the expected use of the datastructure, an O(n) cleanup phase might be appropriate, or you could just throw an exception.
Another issue that might need to be considered is multi-threading. Three obvious problems:
- if
id++
isn't atomic and two threads obtained a newid
at the same time then you could get two entries with the same id. - if the incrementation of id and the assignment of the created tuple aren't atomic together (they're probably not) and there were two simultaneous inserts, then you could get a race condition where the older id wins.
- if the assignment of the tuple isn't atomic, and there's an
insert()
at the same time as aget()
then you might end up in a position where you've got saynew_id, old_value
in the array, causing the wrong value to be returned.
If any of these are problems, the absolute easiest solution to this is to put "not thread safe" in the documentation (cheating). Alternatively, if you can't implement the methods atomically in your language of choice, you'd need to put some sort of synchronisation locks around them.
How about an array
of tuples timestamp, value
, with an additional timestamp, value
called all
. Since you only care about the relative time of insertions, you can use a monotonically increasing id
for the values of timestamp:
type Entry
int timestamp,
int value
type structure
int id
Entry all
Entry array
Initialise all fields to 0. Then the following should work for you:
setValue(index i, value v):
array[i] = id++, value
value getValue(index i)
if(all.timestamp > array[i].timestamp) return all.value
else return array[i].valuesetAll(value v)
all = id++, value
A problem with this approach is that eventually you'll run out of ids for timestamp, and might wrap around. If you chose a 64 bit value to store timestamps, then this gives you 18,446,744,073,709,551,616 insertions or setAlls before this happens. Depending on the expected use of the datastructure, an O(n) cleanup phase might be appropriate, or you could just throw an exception.
Another issue that might need to be considered is multi-threading. Three obvious problems:
- if
id++
isn't atomic and two threads obtained a newid
at the same time then you could get two entries with the same id. - if the incrementation of id and the assignment of the created tuple aren't atomic together (they're probably not) and there were two simultaneous inserts, then you could get a race condition where the older id wins.
- if the assignment of the tuple isn't atomic, and there's an
insert()
at the same time as aget()
then you might end up in a position where you've got saynew_id, old_value
in the array, causing the wrong value to be returned.
If any of these are problems, the absolute easiest solution to this is to put "not thread safe" in the documentation (cheating). Alternatively, if you can't implement the methods atomically in your language of choice, you'd need to put some sort of synchronisation locks around them.
edited Dec 19 '17 at 7:05
answered Apr 4 '12 at 6:03
Timothy JonesTimothy Jones
16.2k44275
16.2k44275
But we can;t be sure that how it works internally. This single statement can be O(n).
– ganesshkumar
Apr 4 '12 at 6:05
1
This relies on every call tocurrent_time()
returning a different value though (otherwise, two timestamps may be equal)
– Damien_The_Unbeliever
Apr 4 '12 at 6:07
@Damien_The_Unbeliever that's true. You may also have to deal with wrap-around cases too, if the program was around for a while.
– Timothy Jones
Apr 4 '12 at 6:15
Thanks Timothy. Perfect!
– Alec
Apr 4 '12 at 6:42
1
Your signature forgetValue()
needs to be fixed.
– Patrick Roberts
Dec 18 '17 at 7:57
|
show 6 more comments
But we can;t be sure that how it works internally. This single statement can be O(n).
– ganesshkumar
Apr 4 '12 at 6:05
1
This relies on every call tocurrent_time()
returning a different value though (otherwise, two timestamps may be equal)
– Damien_The_Unbeliever
Apr 4 '12 at 6:07
@Damien_The_Unbeliever that's true. You may also have to deal with wrap-around cases too, if the program was around for a while.
– Timothy Jones
Apr 4 '12 at 6:15
Thanks Timothy. Perfect!
– Alec
Apr 4 '12 at 6:42
1
Your signature forgetValue()
needs to be fixed.
– Patrick Roberts
Dec 18 '17 at 7:57
But we can;t be sure that how it works internally. This single statement can be O(n).
– ganesshkumar
Apr 4 '12 at 6:05
But we can;t be sure that how it works internally. This single statement can be O(n).
– ganesshkumar
Apr 4 '12 at 6:05
1
1
This relies on every call to
current_time()
returning a different value though (otherwise, two timestamps may be equal)– Damien_The_Unbeliever
Apr 4 '12 at 6:07
This relies on every call to
current_time()
returning a different value though (otherwise, two timestamps may be equal)– Damien_The_Unbeliever
Apr 4 '12 at 6:07
@Damien_The_Unbeliever that's true. You may also have to deal with wrap-around cases too, if the program was around for a while.
– Timothy Jones
Apr 4 '12 at 6:15
@Damien_The_Unbeliever that's true. You may also have to deal with wrap-around cases too, if the program was around for a while.
– Timothy Jones
Apr 4 '12 at 6:15
Thanks Timothy. Perfect!
– Alec
Apr 4 '12 at 6:42
Thanks Timothy. Perfect!
– Alec
Apr 4 '12 at 6:42
1
1
Your signature for
getValue()
needs to be fixed.– Patrick Roberts
Dec 18 '17 at 7:57
Your signature for
getValue()
needs to be fixed.– Patrick Roberts
Dec 18 '17 at 7:57
|
show 6 more comments
I got the same question in one of the technical interviews.
Here is my complete ready-to-use Java-implementation including the test cases.
The key idea is to keep the value of setAll()
in special variable (e.g. joker
) and then trace the change of this value in a proper way.
In order to keep the code clean, some access modifiers have been abolished.
Node class:
import java.time.LocalDate;
class Node
int value;
LocalDate jokerSetTime;
Node(int val, LocalDate jokSetTime)
value = val;
jokerSetTime = jokSetTime;
DS class:
class DS
Node arr;
DS(int len)
arr = new Node[len];
DataStructure class:
import java.time.LocalDate;
class DataStructure
private boolean isJokerChanged;
private Integer joker;
private LocalDate jokerSetTime;
private DS myDS;
DataStructure(int len)
myDS = new DS(len);
Integer get(int i)
Integer result;
if (myDS.arr.length < i)
return null;
// setAll() has been just called
if (isJokerChanged)
return joker;
if (myDS.arr[i] == null)
// setAll() has been previously called
if (joker != null)
result = joker;
else
result = null;
else if (myDS.arr[i].jokerSetTime == jokerSetTime)
// cell value has been overridden after setAll() call
result = myDS.arr[i].value;
else
result = joker;
return result;
void setAll(int val)
isJokerChanged = true;
joker = val;
jokerSetTime = LocalDate.now();
void set(int i, int val)
isJokerChanged = false;
myDS.arr[i] = new Node(val, jokerSetTime);
Main class:
class Main
public static void main(String args)
DataStructure ds = new DataStructure(100);
Integer res;
res = ds.get(3);
ds.set(3, 10);
res = ds.get(3);
ds.setAll(6);
res = ds.get(3);
res = ds.get(15);
ds.set(4, 7);
res = ds.get(4);
res = ds.get(3);
ds.setAll(6);
ds.set(8, 2);
res = ds.get(3);
Update:
The code has been updated. The previous implementation didn't take into account the case when setAll()
is called twice in a row with the same value and is followed by set(x)
, get(y)
, e.g.: setAll(100)
, set(3, 1)
, setAll(100)
, set(5, 3)
, get(3)
.
The use of timestamp approach has been added to allow distinguishing between different setAll()
calls with identical values.
P.S. This implementation is not a thread safe.
2
I have a strong hunch that this was expected. All the timestamp based solutions add completely unnecessary complexity.
– user4815162342
Apr 10 '18 at 14:15
2
@MikeB. Great implementation! Very clean and clear. At first I was having trouble understanding the previous solutions with all of the timestamps, but this is just amazing, exactly what I was looking for. Thanks again!
– Martin
Jun 10 '18 at 9:54
1
This solution is wrong. Consider the following example in the main function:ds.setAll(6); ds.set(3,10); ds.setAll(6); ds.set(4,1); res = ds.get();
It will return the incorrect value of 10, even thoughsetAll(6)
was called after it. The secondds.set()
call is needed to clear theisJokerChanged
variable, which would have superficially defended against this bug otherwise.
– AnatolyVorobey
Jul 12 '18 at 17:29
@AnatolyVorobey, the code has been fixed. Thanks for the test case.
– Mike B.
Jul 15 '18 at 12:34
add a comment |
I got the same question in one of the technical interviews.
Here is my complete ready-to-use Java-implementation including the test cases.
The key idea is to keep the value of setAll()
in special variable (e.g. joker
) and then trace the change of this value in a proper way.
In order to keep the code clean, some access modifiers have been abolished.
Node class:
import java.time.LocalDate;
class Node
int value;
LocalDate jokerSetTime;
Node(int val, LocalDate jokSetTime)
value = val;
jokerSetTime = jokSetTime;
DS class:
class DS
Node arr;
DS(int len)
arr = new Node[len];
DataStructure class:
import java.time.LocalDate;
class DataStructure
private boolean isJokerChanged;
private Integer joker;
private LocalDate jokerSetTime;
private DS myDS;
DataStructure(int len)
myDS = new DS(len);
Integer get(int i)
Integer result;
if (myDS.arr.length < i)
return null;
// setAll() has been just called
if (isJokerChanged)
return joker;
if (myDS.arr[i] == null)
// setAll() has been previously called
if (joker != null)
result = joker;
else
result = null;
else if (myDS.arr[i].jokerSetTime == jokerSetTime)
// cell value has been overridden after setAll() call
result = myDS.arr[i].value;
else
result = joker;
return result;
void setAll(int val)
isJokerChanged = true;
joker = val;
jokerSetTime = LocalDate.now();
void set(int i, int val)
isJokerChanged = false;
myDS.arr[i] = new Node(val, jokerSetTime);
Main class:
class Main
public static void main(String args)
DataStructure ds = new DataStructure(100);
Integer res;
res = ds.get(3);
ds.set(3, 10);
res = ds.get(3);
ds.setAll(6);
res = ds.get(3);
res = ds.get(15);
ds.set(4, 7);
res = ds.get(4);
res = ds.get(3);
ds.setAll(6);
ds.set(8, 2);
res = ds.get(3);
Update:
The code has been updated. The previous implementation didn't take into account the case when setAll()
is called twice in a row with the same value and is followed by set(x)
, get(y)
, e.g.: setAll(100)
, set(3, 1)
, setAll(100)
, set(5, 3)
, get(3)
.
The use of timestamp approach has been added to allow distinguishing between different setAll()
calls with identical values.
P.S. This implementation is not a thread safe.
2
I have a strong hunch that this was expected. All the timestamp based solutions add completely unnecessary complexity.
– user4815162342
Apr 10 '18 at 14:15
2
@MikeB. Great implementation! Very clean and clear. At first I was having trouble understanding the previous solutions with all of the timestamps, but this is just amazing, exactly what I was looking for. Thanks again!
– Martin
Jun 10 '18 at 9:54
1
This solution is wrong. Consider the following example in the main function:ds.setAll(6); ds.set(3,10); ds.setAll(6); ds.set(4,1); res = ds.get();
It will return the incorrect value of 10, even thoughsetAll(6)
was called after it. The secondds.set()
call is needed to clear theisJokerChanged
variable, which would have superficially defended against this bug otherwise.
– AnatolyVorobey
Jul 12 '18 at 17:29
@AnatolyVorobey, the code has been fixed. Thanks for the test case.
– Mike B.
Jul 15 '18 at 12:34
add a comment |
I got the same question in one of the technical interviews.
Here is my complete ready-to-use Java-implementation including the test cases.
The key idea is to keep the value of setAll()
in special variable (e.g. joker
) and then trace the change of this value in a proper way.
In order to keep the code clean, some access modifiers have been abolished.
Node class:
import java.time.LocalDate;
class Node
int value;
LocalDate jokerSetTime;
Node(int val, LocalDate jokSetTime)
value = val;
jokerSetTime = jokSetTime;
DS class:
class DS
Node arr;
DS(int len)
arr = new Node[len];
DataStructure class:
import java.time.LocalDate;
class DataStructure
private boolean isJokerChanged;
private Integer joker;
private LocalDate jokerSetTime;
private DS myDS;
DataStructure(int len)
myDS = new DS(len);
Integer get(int i)
Integer result;
if (myDS.arr.length < i)
return null;
// setAll() has been just called
if (isJokerChanged)
return joker;
if (myDS.arr[i] == null)
// setAll() has been previously called
if (joker != null)
result = joker;
else
result = null;
else if (myDS.arr[i].jokerSetTime == jokerSetTime)
// cell value has been overridden after setAll() call
result = myDS.arr[i].value;
else
result = joker;
return result;
void setAll(int val)
isJokerChanged = true;
joker = val;
jokerSetTime = LocalDate.now();
void set(int i, int val)
isJokerChanged = false;
myDS.arr[i] = new Node(val, jokerSetTime);
Main class:
class Main
public static void main(String args)
DataStructure ds = new DataStructure(100);
Integer res;
res = ds.get(3);
ds.set(3, 10);
res = ds.get(3);
ds.setAll(6);
res = ds.get(3);
res = ds.get(15);
ds.set(4, 7);
res = ds.get(4);
res = ds.get(3);
ds.setAll(6);
ds.set(8, 2);
res = ds.get(3);
Update:
The code has been updated. The previous implementation didn't take into account the case when setAll()
is called twice in a row with the same value and is followed by set(x)
, get(y)
, e.g.: setAll(100)
, set(3, 1)
, setAll(100)
, set(5, 3)
, get(3)
.
The use of timestamp approach has been added to allow distinguishing between different setAll()
calls with identical values.
P.S. This implementation is not a thread safe.
I got the same question in one of the technical interviews.
Here is my complete ready-to-use Java-implementation including the test cases.
The key idea is to keep the value of setAll()
in special variable (e.g. joker
) and then trace the change of this value in a proper way.
In order to keep the code clean, some access modifiers have been abolished.
Node class:
import java.time.LocalDate;
class Node
int value;
LocalDate jokerSetTime;
Node(int val, LocalDate jokSetTime)
value = val;
jokerSetTime = jokSetTime;
DS class:
class DS
Node arr;
DS(int len)
arr = new Node[len];
DataStructure class:
import java.time.LocalDate;
class DataStructure
private boolean isJokerChanged;
private Integer joker;
private LocalDate jokerSetTime;
private DS myDS;
DataStructure(int len)
myDS = new DS(len);
Integer get(int i)
Integer result;
if (myDS.arr.length < i)
return null;
// setAll() has been just called
if (isJokerChanged)
return joker;
if (myDS.arr[i] == null)
// setAll() has been previously called
if (joker != null)
result = joker;
else
result = null;
else if (myDS.arr[i].jokerSetTime == jokerSetTime)
// cell value has been overridden after setAll() call
result = myDS.arr[i].value;
else
result = joker;
return result;
void setAll(int val)
isJokerChanged = true;
joker = val;
jokerSetTime = LocalDate.now();
void set(int i, int val)
isJokerChanged = false;
myDS.arr[i] = new Node(val, jokerSetTime);
Main class:
class Main
public static void main(String args)
DataStructure ds = new DataStructure(100);
Integer res;
res = ds.get(3);
ds.set(3, 10);
res = ds.get(3);
ds.setAll(6);
res = ds.get(3);
res = ds.get(15);
ds.set(4, 7);
res = ds.get(4);
res = ds.get(3);
ds.setAll(6);
ds.set(8, 2);
res = ds.get(3);
Update:
The code has been updated. The previous implementation didn't take into account the case when setAll()
is called twice in a row with the same value and is followed by set(x)
, get(y)
, e.g.: setAll(100)
, set(3, 1)
, setAll(100)
, set(5, 3)
, get(3)
.
The use of timestamp approach has been added to allow distinguishing between different setAll()
calls with identical values.
P.S. This implementation is not a thread safe.
edited Jul 15 '18 at 14:27
answered Jan 17 '17 at 7:17
Mike B.Mike B.
4,758135685
4,758135685
2
I have a strong hunch that this was expected. All the timestamp based solutions add completely unnecessary complexity.
– user4815162342
Apr 10 '18 at 14:15
2
@MikeB. Great implementation! Very clean and clear. At first I was having trouble understanding the previous solutions with all of the timestamps, but this is just amazing, exactly what I was looking for. Thanks again!
– Martin
Jun 10 '18 at 9:54
1
This solution is wrong. Consider the following example in the main function:ds.setAll(6); ds.set(3,10); ds.setAll(6); ds.set(4,1); res = ds.get();
It will return the incorrect value of 10, even thoughsetAll(6)
was called after it. The secondds.set()
call is needed to clear theisJokerChanged
variable, which would have superficially defended against this bug otherwise.
– AnatolyVorobey
Jul 12 '18 at 17:29
@AnatolyVorobey, the code has been fixed. Thanks for the test case.
– Mike B.
Jul 15 '18 at 12:34
add a comment |
2
I have a strong hunch that this was expected. All the timestamp based solutions add completely unnecessary complexity.
– user4815162342
Apr 10 '18 at 14:15
2
@MikeB. Great implementation! Very clean and clear. At first I was having trouble understanding the previous solutions with all of the timestamps, but this is just amazing, exactly what I was looking for. Thanks again!
– Martin
Jun 10 '18 at 9:54
1
This solution is wrong. Consider the following example in the main function:ds.setAll(6); ds.set(3,10); ds.setAll(6); ds.set(4,1); res = ds.get();
It will return the incorrect value of 10, even thoughsetAll(6)
was called after it. The secondds.set()
call is needed to clear theisJokerChanged
variable, which would have superficially defended against this bug otherwise.
– AnatolyVorobey
Jul 12 '18 at 17:29
@AnatolyVorobey, the code has been fixed. Thanks for the test case.
– Mike B.
Jul 15 '18 at 12:34
2
2
I have a strong hunch that this was expected. All the timestamp based solutions add completely unnecessary complexity.
– user4815162342
Apr 10 '18 at 14:15
I have a strong hunch that this was expected. All the timestamp based solutions add completely unnecessary complexity.
– user4815162342
Apr 10 '18 at 14:15
2
2
@MikeB. Great implementation! Very clean and clear. At first I was having trouble understanding the previous solutions with all of the timestamps, but this is just amazing, exactly what I was looking for. Thanks again!
– Martin
Jun 10 '18 at 9:54
@MikeB. Great implementation! Very clean and clear. At first I was having trouble understanding the previous solutions with all of the timestamps, but this is just amazing, exactly what I was looking for. Thanks again!
– Martin
Jun 10 '18 at 9:54
1
1
This solution is wrong. Consider the following example in the main function:
ds.setAll(6); ds.set(3,10); ds.setAll(6); ds.set(4,1); res = ds.get();
It will return the incorrect value of 10, even though setAll(6)
was called after it. The second ds.set()
call is needed to clear the isJokerChanged
variable, which would have superficially defended against this bug otherwise.– AnatolyVorobey
Jul 12 '18 at 17:29
This solution is wrong. Consider the following example in the main function:
ds.setAll(6); ds.set(3,10); ds.setAll(6); ds.set(4,1); res = ds.get();
It will return the incorrect value of 10, even though setAll(6)
was called after it. The second ds.set()
call is needed to clear the isJokerChanged
variable, which would have superficially defended against this bug otherwise.– AnatolyVorobey
Jul 12 '18 at 17:29
@AnatolyVorobey, the code has been fixed. Thanks for the test case.
– Mike B.
Jul 15 '18 at 12:34
@AnatolyVorobey, the code has been fixed. Thanks for the test case.
– Mike B.
Jul 15 '18 at 12:34
add a comment |
How about an array of pointers to a single common value? Set the value, and all references will point to the single changed value in O(1)..
Answers the question though! ;) lol.
– WOPR
Apr 4 '12 at 6:07
This works until you want to change one of the elements to point to something other than the common value.
– Timothy Jones
Apr 4 '12 at 6:09
That wasn't a requirement specified in the question.
– WOPR
Apr 4 '12 at 6:10
1
I think it is - if you call setAll(), and then setValue(), you'd have to change one of the elements.
– Timothy Jones
Apr 4 '12 at 6:16
3
I would say that specs clearly imply that it should be possible to set values individually without affecting the other values. Otherwise the index parameter to the setValue method is superfluous. And if you do not want to interpret method signatures as specifications, then it would be possible to simply implement all of the methods as NoOps...
– Alderath
Apr 4 '12 at 8:57
add a comment |
How about an array of pointers to a single common value? Set the value, and all references will point to the single changed value in O(1)..
Answers the question though! ;) lol.
– WOPR
Apr 4 '12 at 6:07
This works until you want to change one of the elements to point to something other than the common value.
– Timothy Jones
Apr 4 '12 at 6:09
That wasn't a requirement specified in the question.
– WOPR
Apr 4 '12 at 6:10
1
I think it is - if you call setAll(), and then setValue(), you'd have to change one of the elements.
– Timothy Jones
Apr 4 '12 at 6:16
3
I would say that specs clearly imply that it should be possible to set values individually without affecting the other values. Otherwise the index parameter to the setValue method is superfluous. And if you do not want to interpret method signatures as specifications, then it would be possible to simply implement all of the methods as NoOps...
– Alderath
Apr 4 '12 at 8:57
add a comment |
How about an array of pointers to a single common value? Set the value, and all references will point to the single changed value in O(1)..
How about an array of pointers to a single common value? Set the value, and all references will point to the single changed value in O(1)..
answered Apr 4 '12 at 6:06
WOPRWOPR
3,73353957
3,73353957
Answers the question though! ;) lol.
– WOPR
Apr 4 '12 at 6:07
This works until you want to change one of the elements to point to something other than the common value.
– Timothy Jones
Apr 4 '12 at 6:09
That wasn't a requirement specified in the question.
– WOPR
Apr 4 '12 at 6:10
1
I think it is - if you call setAll(), and then setValue(), you'd have to change one of the elements.
– Timothy Jones
Apr 4 '12 at 6:16
3
I would say that specs clearly imply that it should be possible to set values individually without affecting the other values. Otherwise the index parameter to the setValue method is superfluous. And if you do not want to interpret method signatures as specifications, then it would be possible to simply implement all of the methods as NoOps...
– Alderath
Apr 4 '12 at 8:57
add a comment |
Answers the question though! ;) lol.
– WOPR
Apr 4 '12 at 6:07
This works until you want to change one of the elements to point to something other than the common value.
– Timothy Jones
Apr 4 '12 at 6:09
That wasn't a requirement specified in the question.
– WOPR
Apr 4 '12 at 6:10
1
I think it is - if you call setAll(), and then setValue(), you'd have to change one of the elements.
– Timothy Jones
Apr 4 '12 at 6:16
3
I would say that specs clearly imply that it should be possible to set values individually without affecting the other values. Otherwise the index parameter to the setValue method is superfluous. And if you do not want to interpret method signatures as specifications, then it would be possible to simply implement all of the methods as NoOps...
– Alderath
Apr 4 '12 at 8:57
Answers the question though! ;) lol.
– WOPR
Apr 4 '12 at 6:07
Answers the question though! ;) lol.
– WOPR
Apr 4 '12 at 6:07
This works until you want to change one of the elements to point to something other than the common value.
– Timothy Jones
Apr 4 '12 at 6:09
This works until you want to change one of the elements to point to something other than the common value.
– Timothy Jones
Apr 4 '12 at 6:09
That wasn't a requirement specified in the question.
– WOPR
Apr 4 '12 at 6:10
That wasn't a requirement specified in the question.
– WOPR
Apr 4 '12 at 6:10
1
1
I think it is - if you call setAll(), and then setValue(), you'd have to change one of the elements.
– Timothy Jones
Apr 4 '12 at 6:16
I think it is - if you call setAll(), and then setValue(), you'd have to change one of the elements.
– Timothy Jones
Apr 4 '12 at 6:16
3
3
I would say that specs clearly imply that it should be possible to set values individually without affecting the other values. Otherwise the index parameter to the setValue method is superfluous. And if you do not want to interpret method signatures as specifications, then it would be possible to simply implement all of the methods as NoOps...
– Alderath
Apr 4 '12 at 8:57
I would say that specs clearly imply that it should be possible to set values individually without affecting the other values. Otherwise the index parameter to the setValue method is superfluous. And if you do not want to interpret method signatures as specifications, then it would be possible to simply implement all of the methods as NoOps...
– Alderath
Apr 4 '12 at 8:57
add a comment |
I was just asked this question very recently in an Interview. I came up with a hash table implementation. It would solve the problem of running out of the timestamp values but the thread safety feature needs to be implemented (probably using lazy initialization techniques)
Let's say in our class we have a private variable _defaultValue to hold a default value and we also have a private hashtable or dictionary _hashtable.
SetAllValues could just set _defaultValue equal to the value passed and _hashtable initialized/set to a new hashtable and discard any reference to the old hash table. SetValue should just add a new value to _hashtable or update the value if the key ( or index) is already present in the _hashtable. GetValue should check whether the key (or index) is present in _hashtable, then return it, else return the value stored in _defaultValue.
This is my first answer on StackOverflow. I am a little lazy in writing up the code. Will probably edit the answer soon.
The interviewer nodded yes to this solution but insisted to implement it without using a hashtable. I guess, he was asking me to give a similar approach as Timothy's answer. And I wasn't able to get it at that moment :(. Anyways, Cheers!
EDIT:
Posting the code (in C#)
class MyDataStruct
private int _defaultValue;
private Dictionary<int,int> _hashtable;
public MyDataStruct()
_defaultValue = 0; // initialize default with some value
_hashtable = new Dictionary<int, int>();
public void SetAllValues(int val)
_defaultValue = val;
_hashtable = new Dictionary<int, int>();
public void SetValue(int index, int val)
if (_hashtable.ContainsKey(index))
_hashtable.Add(index, val);
else
_hashtable[index] = val;
public int GetValue(int index)
if (_hashtable.ContainsKey(index))
return _hashtable[index];
else
return _defaultValue;
could you please provide your code with hash table?
– Hengameh
May 2 '15 at 6:07
1
@Hengameh -- added!
– tapas nayak
Jul 20 '15 at 6:30
1
This is almost correct except that GetValue for a non existing hey would always return default value even when this key was not inserted to the data structure.
– akrabi
Nov 18 '15 at 18:11
add a comment |
I was just asked this question very recently in an Interview. I came up with a hash table implementation. It would solve the problem of running out of the timestamp values but the thread safety feature needs to be implemented (probably using lazy initialization techniques)
Let's say in our class we have a private variable _defaultValue to hold a default value and we also have a private hashtable or dictionary _hashtable.
SetAllValues could just set _defaultValue equal to the value passed and _hashtable initialized/set to a new hashtable and discard any reference to the old hash table. SetValue should just add a new value to _hashtable or update the value if the key ( or index) is already present in the _hashtable. GetValue should check whether the key (or index) is present in _hashtable, then return it, else return the value stored in _defaultValue.
This is my first answer on StackOverflow. I am a little lazy in writing up the code. Will probably edit the answer soon.
The interviewer nodded yes to this solution but insisted to implement it without using a hashtable. I guess, he was asking me to give a similar approach as Timothy's answer. And I wasn't able to get it at that moment :(. Anyways, Cheers!
EDIT:
Posting the code (in C#)
class MyDataStruct
private int _defaultValue;
private Dictionary<int,int> _hashtable;
public MyDataStruct()
_defaultValue = 0; // initialize default with some value
_hashtable = new Dictionary<int, int>();
public void SetAllValues(int val)
_defaultValue = val;
_hashtable = new Dictionary<int, int>();
public void SetValue(int index, int val)
if (_hashtable.ContainsKey(index))
_hashtable.Add(index, val);
else
_hashtable[index] = val;
public int GetValue(int index)
if (_hashtable.ContainsKey(index))
return _hashtable[index];
else
return _defaultValue;
could you please provide your code with hash table?
– Hengameh
May 2 '15 at 6:07
1
@Hengameh -- added!
– tapas nayak
Jul 20 '15 at 6:30
1
This is almost correct except that GetValue for a non existing hey would always return default value even when this key was not inserted to the data structure.
– akrabi
Nov 18 '15 at 18:11
add a comment |
I was just asked this question very recently in an Interview. I came up with a hash table implementation. It would solve the problem of running out of the timestamp values but the thread safety feature needs to be implemented (probably using lazy initialization techniques)
Let's say in our class we have a private variable _defaultValue to hold a default value and we also have a private hashtable or dictionary _hashtable.
SetAllValues could just set _defaultValue equal to the value passed and _hashtable initialized/set to a new hashtable and discard any reference to the old hash table. SetValue should just add a new value to _hashtable or update the value if the key ( or index) is already present in the _hashtable. GetValue should check whether the key (or index) is present in _hashtable, then return it, else return the value stored in _defaultValue.
This is my first answer on StackOverflow. I am a little lazy in writing up the code. Will probably edit the answer soon.
The interviewer nodded yes to this solution but insisted to implement it without using a hashtable. I guess, he was asking me to give a similar approach as Timothy's answer. And I wasn't able to get it at that moment :(. Anyways, Cheers!
EDIT:
Posting the code (in C#)
class MyDataStruct
private int _defaultValue;
private Dictionary<int,int> _hashtable;
public MyDataStruct()
_defaultValue = 0; // initialize default with some value
_hashtable = new Dictionary<int, int>();
public void SetAllValues(int val)
_defaultValue = val;
_hashtable = new Dictionary<int, int>();
public void SetValue(int index, int val)
if (_hashtable.ContainsKey(index))
_hashtable.Add(index, val);
else
_hashtable[index] = val;
public int GetValue(int index)
if (_hashtable.ContainsKey(index))
return _hashtable[index];
else
return _defaultValue;
I was just asked this question very recently in an Interview. I came up with a hash table implementation. It would solve the problem of running out of the timestamp values but the thread safety feature needs to be implemented (probably using lazy initialization techniques)
Let's say in our class we have a private variable _defaultValue to hold a default value and we also have a private hashtable or dictionary _hashtable.
SetAllValues could just set _defaultValue equal to the value passed and _hashtable initialized/set to a new hashtable and discard any reference to the old hash table. SetValue should just add a new value to _hashtable or update the value if the key ( or index) is already present in the _hashtable. GetValue should check whether the key (or index) is present in _hashtable, then return it, else return the value stored in _defaultValue.
This is my first answer on StackOverflow. I am a little lazy in writing up the code. Will probably edit the answer soon.
The interviewer nodded yes to this solution but insisted to implement it without using a hashtable. I guess, he was asking me to give a similar approach as Timothy's answer. And I wasn't able to get it at that moment :(. Anyways, Cheers!
EDIT:
Posting the code (in C#)
class MyDataStruct
private int _defaultValue;
private Dictionary<int,int> _hashtable;
public MyDataStruct()
_defaultValue = 0; // initialize default with some value
_hashtable = new Dictionary<int, int>();
public void SetAllValues(int val)
_defaultValue = val;
_hashtable = new Dictionary<int, int>();
public void SetValue(int index, int val)
if (_hashtable.ContainsKey(index))
_hashtable.Add(index, val);
else
_hashtable[index] = val;
public int GetValue(int index)
if (_hashtable.ContainsKey(index))
return _hashtable[index];
else
return _defaultValue;
edited Dec 18 '17 at 7:29
eFarzad
564721
564721
answered Aug 23 '14 at 7:24
tapas nayaktapas nayak
313
313
could you please provide your code with hash table?
– Hengameh
May 2 '15 at 6:07
1
@Hengameh -- added!
– tapas nayak
Jul 20 '15 at 6:30
1
This is almost correct except that GetValue for a non existing hey would always return default value even when this key was not inserted to the data structure.
– akrabi
Nov 18 '15 at 18:11
add a comment |
could you please provide your code with hash table?
– Hengameh
May 2 '15 at 6:07
1
@Hengameh -- added!
– tapas nayak
Jul 20 '15 at 6:30
1
This is almost correct except that GetValue for a non existing hey would always return default value even when this key was not inserted to the data structure.
– akrabi
Nov 18 '15 at 18:11
could you please provide your code with hash table?
– Hengameh
May 2 '15 at 6:07
could you please provide your code with hash table?
– Hengameh
May 2 '15 at 6:07
1
1
@Hengameh -- added!
– tapas nayak
Jul 20 '15 at 6:30
@Hengameh -- added!
– tapas nayak
Jul 20 '15 at 6:30
1
1
This is almost correct except that GetValue for a non existing hey would always return default value even when this key was not inserted to the data structure.
– akrabi
Nov 18 '15 at 18:11
This is almost correct except that GetValue for a non existing hey would always return default value even when this key was not inserted to the data structure.
– akrabi
Nov 18 '15 at 18:11
add a comment |
We can have a variable V which stores an int and an array of containing Tuple as Value, id..
And a global int variable G (which will act like identity in SQL and whenever any set or setAll operation is done its value get increment by 1)
initial all Ids and V value will be default say null..
so V = null All Tuple = null, null
set(index i, val v) -> G = G+1, arr[i].Val = v, arr[i].Id = G
get(index i) -> if V.Id > arr[i].Id return V.Val else return arr[i]
set-all(val v) -> G= G+1, V.Id= G, V.Val = v
add a comment |
We can have a variable V which stores an int and an array of containing Tuple as Value, id..
And a global int variable G (which will act like identity in SQL and whenever any set or setAll operation is done its value get increment by 1)
initial all Ids and V value will be default say null..
so V = null All Tuple = null, null
set(index i, val v) -> G = G+1, arr[i].Val = v, arr[i].Id = G
get(index i) -> if V.Id > arr[i].Id return V.Val else return arr[i]
set-all(val v) -> G= G+1, V.Id= G, V.Val = v
add a comment |
We can have a variable V which stores an int and an array of containing Tuple as Value, id..
And a global int variable G (which will act like identity in SQL and whenever any set or setAll operation is done its value get increment by 1)
initial all Ids and V value will be default say null..
so V = null All Tuple = null, null
set(index i, val v) -> G = G+1, arr[i].Val = v, arr[i].Id = G
get(index i) -> if V.Id > arr[i].Id return V.Val else return arr[i]
set-all(val v) -> G= G+1, V.Id= G, V.Val = v
We can have a variable V which stores an int and an array of containing Tuple as Value, id..
And a global int variable G (which will act like identity in SQL and whenever any set or setAll operation is done its value get increment by 1)
initial all Ids and V value will be default say null..
so V = null All Tuple = null, null
set(index i, val v) -> G = G+1, arr[i].Val = v, arr[i].Id = G
get(index i) -> if V.Id > arr[i].Id return V.Val else return arr[i]
set-all(val v) -> G= G+1, V.Id= G, V.Val = v
edited Aug 23 '14 at 15:24
VMai
8,85261631
8,85261631
answered Aug 23 '14 at 8:11
Pramod GuptaPramod Gupta
263
263
add a comment |
add a comment |
All existing answers use a timestamp that is incremented on every setVal
operation. This is not necessary. In fact, it is necessary only to increment the timestamp on setAll
. Another issue some raised was arithmetic overflow. This can be handled without breaking the constant time bounds by updating a single cell on each setAll
and performing the time comparison carefully.
How it works
The basic concept is essentially similar to that of the other answers, but with a twist.
What they do: Store the value used for the last setAll
operation separately, and keep track of the time that operation was performed. Each time they perform a setVal
, they store the current time along with the given value in the array. Each time they perform a getVal
, they compare the time in the given location to the time the last setAll
occurred, and then choose either the value at the location or the setAll
value depending on which is greater.
Why this can be a problem: Suppose the current time overflows and a setAll
operation occurs soon afterward. It will appear as though most of the stored array values are newer than the setAll
value, when they are actually older.
The solution: Stop imagining that we're keeping track of the total amount of time that has passed since data structure initialization. Imagine a giant clock with a "second hand" that ticks not 60 times around the circle but rather 2^n times around the circle. The position of the second hand represents the time of the most recent setAll
operation. Each setVal
operation stores this time along with a value. So if we perform a setAll
when the "clock" is at 45, and then perform six setVal
operations on different elements, the setAll
time and the times at all six locations will be the same. We wish to maintain the following invariant:
The time at a given element location equals the
setAll
time if and only if that element was set withsetVal
more recently than the lastsetAll
operation.
Clearly, the procedure described above automatically ensures that if the element was set recently, then its time will equal the setAll
time. The challenge is to make the reverse implication hold as well.
To be continued ....
The code
I've written this in Haskell because that is the language I know best, but it is not exactly the most natural language for the job.
-# LANGUAGE BangPatterns #-
module RepArr where
import Control.Monad.Primitive
import Data.Primitive.MutVar
import qualified Data.Vector.Mutable as V
import Data.Vector.Mutable (MVector)
import Control.Applicative
import Prelude hiding (replicate)
import Control.Monad.ST
import Data.Word
-- The Int in the MutVar is the refresh pointer
data RepArr s a = RepArr (MutVar s (Word, a, Int)) (MVector s (Word,a))
-- Create a fancy array of a given length, initially filled with a given value
replicate :: (PrimMonad m, Applicative m) => Int -> a -> m (RepArr (PrimState m) a)
replicate n a = RepArr <$> newMutVar (0,a,0) <*> V.replicate n (0,a)
getVal :: PrimMonad m => RepArr (PrimState m) a -> Int -> m a
getVal (RepArr allv vec) n = do
(vectime, vecval) <- V.read vec n
(alltime, allval, _) <- readMutVar allv
if (fromIntegral (alltime - vectime) :: Int) > 0
then return allval
else return vecval
setVal :: PrimMonad m => RepArr (PrimState m) a -> Int -> a -> m ()
setVal (RepArr allv vec) i v = do
(!alltime, _, _) <- readMutVar allv
V.write vec i (alltime, v)
setAll :: PrimMonad m => RepArr (PrimState m) a -> a -> m ()
setAll r@(RepArr allv vec) v = do
(oldt, _, oldc) <- readMutVar allv
getVal r oldc >>= setVal r oldc
let !newc = case oldc+1 of
op1 | op1 == V.length vec -> 0
| otherwise -> op1
let !newt = oldt+1
writeMutVar allv (newt, v, newc)
To avoid potential (rare) garbage collection pauses, it's actually necessary to unbox the Int
and Word
values, as well as using unboxed vectors instead of polymorphic ones, but I'm not really in the mood and this is a fairly mechanical task.
Here's a version in C (completely untested):
#include <malloc.h>
struct Pair
unsigned long time;
void* val;
;
struct RepArr
unsigned long allT;
void* allV;
long count;
long length;
struct Pair vec;
;
struct RepArr *replicate (long n, void* val)
struct RepArr *q = malloc (sizeof (struct RepArr)+n*sizeof (struct Pair));
q->allT = 1;
q->allV = val;
q->count = 0;
q->length = n;
int i;
for (i=0; i<n; i++)
struct Pair foo = 0,val;
q->vec[i] = foo;
return q;
void* getVal (struct RepArr *q, long i)
if ((long)(q->vec[i].time - q->allT) < 0)
return q->allV;
else
return q->vec[i].val;
void setVal (struct RepArr *q, long i, void* val)
q->vec[i].time = q->allT;
q->vec[i].val = val;
void setAll (struct RepArr *q, void* val)
setVal (q, q->count, getVal (q, q->count));
q->allV = val;
q->allT++;
q->count++;
if (q->count == q->length)
q->count = 0;
add a comment |
All existing answers use a timestamp that is incremented on every setVal
operation. This is not necessary. In fact, it is necessary only to increment the timestamp on setAll
. Another issue some raised was arithmetic overflow. This can be handled without breaking the constant time bounds by updating a single cell on each setAll
and performing the time comparison carefully.
How it works
The basic concept is essentially similar to that of the other answers, but with a twist.
What they do: Store the value used for the last setAll
operation separately, and keep track of the time that operation was performed. Each time they perform a setVal
, they store the current time along with the given value in the array. Each time they perform a getVal
, they compare the time in the given location to the time the last setAll
occurred, and then choose either the value at the location or the setAll
value depending on which is greater.
Why this can be a problem: Suppose the current time overflows and a setAll
operation occurs soon afterward. It will appear as though most of the stored array values are newer than the setAll
value, when they are actually older.
The solution: Stop imagining that we're keeping track of the total amount of time that has passed since data structure initialization. Imagine a giant clock with a "second hand" that ticks not 60 times around the circle but rather 2^n times around the circle. The position of the second hand represents the time of the most recent setAll
operation. Each setVal
operation stores this time along with a value. So if we perform a setAll
when the "clock" is at 45, and then perform six setVal
operations on different elements, the setAll
time and the times at all six locations will be the same. We wish to maintain the following invariant:
The time at a given element location equals the
setAll
time if and only if that element was set withsetVal
more recently than the lastsetAll
operation.
Clearly, the procedure described above automatically ensures that if the element was set recently, then its time will equal the setAll
time. The challenge is to make the reverse implication hold as well.
To be continued ....
The code
I've written this in Haskell because that is the language I know best, but it is not exactly the most natural language for the job.
-# LANGUAGE BangPatterns #-
module RepArr where
import Control.Monad.Primitive
import Data.Primitive.MutVar
import qualified Data.Vector.Mutable as V
import Data.Vector.Mutable (MVector)
import Control.Applicative
import Prelude hiding (replicate)
import Control.Monad.ST
import Data.Word
-- The Int in the MutVar is the refresh pointer
data RepArr s a = RepArr (MutVar s (Word, a, Int)) (MVector s (Word,a))
-- Create a fancy array of a given length, initially filled with a given value
replicate :: (PrimMonad m, Applicative m) => Int -> a -> m (RepArr (PrimState m) a)
replicate n a = RepArr <$> newMutVar (0,a,0) <*> V.replicate n (0,a)
getVal :: PrimMonad m => RepArr (PrimState m) a -> Int -> m a
getVal (RepArr allv vec) n = do
(vectime, vecval) <- V.read vec n
(alltime, allval, _) <- readMutVar allv
if (fromIntegral (alltime - vectime) :: Int) > 0
then return allval
else return vecval
setVal :: PrimMonad m => RepArr (PrimState m) a -> Int -> a -> m ()
setVal (RepArr allv vec) i v = do
(!alltime, _, _) <- readMutVar allv
V.write vec i (alltime, v)
setAll :: PrimMonad m => RepArr (PrimState m) a -> a -> m ()
setAll r@(RepArr allv vec) v = do
(oldt, _, oldc) <- readMutVar allv
getVal r oldc >>= setVal r oldc
let !newc = case oldc+1 of
op1 | op1 == V.length vec -> 0
| otherwise -> op1
let !newt = oldt+1
writeMutVar allv (newt, v, newc)
To avoid potential (rare) garbage collection pauses, it's actually necessary to unbox the Int
and Word
values, as well as using unboxed vectors instead of polymorphic ones, but I'm not really in the mood and this is a fairly mechanical task.
Here's a version in C (completely untested):
#include <malloc.h>
struct Pair
unsigned long time;
void* val;
;
struct RepArr
unsigned long allT;
void* allV;
long count;
long length;
struct Pair vec;
;
struct RepArr *replicate (long n, void* val)
struct RepArr *q = malloc (sizeof (struct RepArr)+n*sizeof (struct Pair));
q->allT = 1;
q->allV = val;
q->count = 0;
q->length = n;
int i;
for (i=0; i<n; i++)
struct Pair foo = 0,val;
q->vec[i] = foo;
return q;
void* getVal (struct RepArr *q, long i)
if ((long)(q->vec[i].time - q->allT) < 0)
return q->allV;
else
return q->vec[i].val;
void setVal (struct RepArr *q, long i, void* val)
q->vec[i].time = q->allT;
q->vec[i].val = val;
void setAll (struct RepArr *q, void* val)
setVal (q, q->count, getVal (q, q->count));
q->allV = val;
q->allT++;
q->count++;
if (q->count == q->length)
q->count = 0;
add a comment |
All existing answers use a timestamp that is incremented on every setVal
operation. This is not necessary. In fact, it is necessary only to increment the timestamp on setAll
. Another issue some raised was arithmetic overflow. This can be handled without breaking the constant time bounds by updating a single cell on each setAll
and performing the time comparison carefully.
How it works
The basic concept is essentially similar to that of the other answers, but with a twist.
What they do: Store the value used for the last setAll
operation separately, and keep track of the time that operation was performed. Each time they perform a setVal
, they store the current time along with the given value in the array. Each time they perform a getVal
, they compare the time in the given location to the time the last setAll
occurred, and then choose either the value at the location or the setAll
value depending on which is greater.
Why this can be a problem: Suppose the current time overflows and a setAll
operation occurs soon afterward. It will appear as though most of the stored array values are newer than the setAll
value, when they are actually older.
The solution: Stop imagining that we're keeping track of the total amount of time that has passed since data structure initialization. Imagine a giant clock with a "second hand" that ticks not 60 times around the circle but rather 2^n times around the circle. The position of the second hand represents the time of the most recent setAll
operation. Each setVal
operation stores this time along with a value. So if we perform a setAll
when the "clock" is at 45, and then perform six setVal
operations on different elements, the setAll
time and the times at all six locations will be the same. We wish to maintain the following invariant:
The time at a given element location equals the
setAll
time if and only if that element was set withsetVal
more recently than the lastsetAll
operation.
Clearly, the procedure described above automatically ensures that if the element was set recently, then its time will equal the setAll
time. The challenge is to make the reverse implication hold as well.
To be continued ....
The code
I've written this in Haskell because that is the language I know best, but it is not exactly the most natural language for the job.
-# LANGUAGE BangPatterns #-
module RepArr where
import Control.Monad.Primitive
import Data.Primitive.MutVar
import qualified Data.Vector.Mutable as V
import Data.Vector.Mutable (MVector)
import Control.Applicative
import Prelude hiding (replicate)
import Control.Monad.ST
import Data.Word
-- The Int in the MutVar is the refresh pointer
data RepArr s a = RepArr (MutVar s (Word, a, Int)) (MVector s (Word,a))
-- Create a fancy array of a given length, initially filled with a given value
replicate :: (PrimMonad m, Applicative m) => Int -> a -> m (RepArr (PrimState m) a)
replicate n a = RepArr <$> newMutVar (0,a,0) <*> V.replicate n (0,a)
getVal :: PrimMonad m => RepArr (PrimState m) a -> Int -> m a
getVal (RepArr allv vec) n = do
(vectime, vecval) <- V.read vec n
(alltime, allval, _) <- readMutVar allv
if (fromIntegral (alltime - vectime) :: Int) > 0
then return allval
else return vecval
setVal :: PrimMonad m => RepArr (PrimState m) a -> Int -> a -> m ()
setVal (RepArr allv vec) i v = do
(!alltime, _, _) <- readMutVar allv
V.write vec i (alltime, v)
setAll :: PrimMonad m => RepArr (PrimState m) a -> a -> m ()
setAll r@(RepArr allv vec) v = do
(oldt, _, oldc) <- readMutVar allv
getVal r oldc >>= setVal r oldc
let !newc = case oldc+1 of
op1 | op1 == V.length vec -> 0
| otherwise -> op1
let !newt = oldt+1
writeMutVar allv (newt, v, newc)
To avoid potential (rare) garbage collection pauses, it's actually necessary to unbox the Int
and Word
values, as well as using unboxed vectors instead of polymorphic ones, but I'm not really in the mood and this is a fairly mechanical task.
Here's a version in C (completely untested):
#include <malloc.h>
struct Pair
unsigned long time;
void* val;
;
struct RepArr
unsigned long allT;
void* allV;
long count;
long length;
struct Pair vec;
;
struct RepArr *replicate (long n, void* val)
struct RepArr *q = malloc (sizeof (struct RepArr)+n*sizeof (struct Pair));
q->allT = 1;
q->allV = val;
q->count = 0;
q->length = n;
int i;
for (i=0; i<n; i++)
struct Pair foo = 0,val;
q->vec[i] = foo;
return q;
void* getVal (struct RepArr *q, long i)
if ((long)(q->vec[i].time - q->allT) < 0)
return q->allV;
else
return q->vec[i].val;
void setVal (struct RepArr *q, long i, void* val)
q->vec[i].time = q->allT;
q->vec[i].val = val;
void setAll (struct RepArr *q, void* val)
setVal (q, q->count, getVal (q, q->count));
q->allV = val;
q->allT++;
q->count++;
if (q->count == q->length)
q->count = 0;
All existing answers use a timestamp that is incremented on every setVal
operation. This is not necessary. In fact, it is necessary only to increment the timestamp on setAll
. Another issue some raised was arithmetic overflow. This can be handled without breaking the constant time bounds by updating a single cell on each setAll
and performing the time comparison carefully.
How it works
The basic concept is essentially similar to that of the other answers, but with a twist.
What they do: Store the value used for the last setAll
operation separately, and keep track of the time that operation was performed. Each time they perform a setVal
, they store the current time along with the given value in the array. Each time they perform a getVal
, they compare the time in the given location to the time the last setAll
occurred, and then choose either the value at the location or the setAll
value depending on which is greater.
Why this can be a problem: Suppose the current time overflows and a setAll
operation occurs soon afterward. It will appear as though most of the stored array values are newer than the setAll
value, when they are actually older.
The solution: Stop imagining that we're keeping track of the total amount of time that has passed since data structure initialization. Imagine a giant clock with a "second hand" that ticks not 60 times around the circle but rather 2^n times around the circle. The position of the second hand represents the time of the most recent setAll
operation. Each setVal
operation stores this time along with a value. So if we perform a setAll
when the "clock" is at 45, and then perform six setVal
operations on different elements, the setAll
time and the times at all six locations will be the same. We wish to maintain the following invariant:
The time at a given element location equals the
setAll
time if and only if that element was set withsetVal
more recently than the lastsetAll
operation.
Clearly, the procedure described above automatically ensures that if the element was set recently, then its time will equal the setAll
time. The challenge is to make the reverse implication hold as well.
To be continued ....
The code
I've written this in Haskell because that is the language I know best, but it is not exactly the most natural language for the job.
-# LANGUAGE BangPatterns #-
module RepArr where
import Control.Monad.Primitive
import Data.Primitive.MutVar
import qualified Data.Vector.Mutable as V
import Data.Vector.Mutable (MVector)
import Control.Applicative
import Prelude hiding (replicate)
import Control.Monad.ST
import Data.Word
-- The Int in the MutVar is the refresh pointer
data RepArr s a = RepArr (MutVar s (Word, a, Int)) (MVector s (Word,a))
-- Create a fancy array of a given length, initially filled with a given value
replicate :: (PrimMonad m, Applicative m) => Int -> a -> m (RepArr (PrimState m) a)
replicate n a = RepArr <$> newMutVar (0,a,0) <*> V.replicate n (0,a)
getVal :: PrimMonad m => RepArr (PrimState m) a -> Int -> m a
getVal (RepArr allv vec) n = do
(vectime, vecval) <- V.read vec n
(alltime, allval, _) <- readMutVar allv
if (fromIntegral (alltime - vectime) :: Int) > 0
then return allval
else return vecval
setVal :: PrimMonad m => RepArr (PrimState m) a -> Int -> a -> m ()
setVal (RepArr allv vec) i v = do
(!alltime, _, _) <- readMutVar allv
V.write vec i (alltime, v)
setAll :: PrimMonad m => RepArr (PrimState m) a -> a -> m ()
setAll r@(RepArr allv vec) v = do
(oldt, _, oldc) <- readMutVar allv
getVal r oldc >>= setVal r oldc
let !newc = case oldc+1 of
op1 | op1 == V.length vec -> 0
| otherwise -> op1
let !newt = oldt+1
writeMutVar allv (newt, v, newc)
To avoid potential (rare) garbage collection pauses, it's actually necessary to unbox the Int
and Word
values, as well as using unboxed vectors instead of polymorphic ones, but I'm not really in the mood and this is a fairly mechanical task.
Here's a version in C (completely untested):
#include <malloc.h>
struct Pair
unsigned long time;
void* val;
;
struct RepArr
unsigned long allT;
void* allV;
long count;
long length;
struct Pair vec;
;
struct RepArr *replicate (long n, void* val)
struct RepArr *q = malloc (sizeof (struct RepArr)+n*sizeof (struct Pair));
q->allT = 1;
q->allV = val;
q->count = 0;
q->length = n;
int i;
for (i=0; i<n; i++)
struct Pair foo = 0,val;
q->vec[i] = foo;
return q;
void* getVal (struct RepArr *q, long i)
if ((long)(q->vec[i].time - q->allT) < 0)
return q->allV;
else
return q->vec[i].val;
void setVal (struct RepArr *q, long i, void* val)
q->vec[i].time = q->allT;
q->vec[i].val = val;
void setAll (struct RepArr *q, void* val)
setVal (q, q->count, getVal (q, q->count));
q->allV = val;
q->allT++;
q->count++;
if (q->count == q->length)
q->count = 0;
edited Mar 25 '15 at 18:59
answered Mar 6 '15 at 1:43
dfeuerdfeuer
32.8k348128
32.8k348128
add a comment |
add a comment |
/*
At the time I am writing this, all of the solutions on this page will double (or
more) the amount of space required to store the array. The following solution
reduces the amount of wasted space from Ω(n) to θ(n/w), where w is the number of
bits in a computer "word". On my machine, that's 64.
This prose in this answer is in C comments, so you can copy-and-paste this
answer verbatim and compile it with your C compiler.
*/
#include <stdbool.h>
#include <stddef.h>
#include <stdint.h>
#include <stdlib.h>
/*
The problem is to support reading and writing values in an array in O(1) time
along with bulk writes in which all values in the array are written at once in
O(1) time. This is possible using a technique invented by Aho, Hopcroft and
Ullman, as far as I know. I will present a version due to Gonzalo Navarro,
"Constant-Time Array Initialization in Little
Space".
The idea is to keep three metadata arrays along with the data array. We also
keep two integers: unset
, which is the last value used in a bulk write
operation, and size
, an approximation for the number of values that have been
set since the last bulk write. At all times, the number of distinct values
written since the last bulk write is between size
and w * size
.
The three metadata arrays describe information about blocks of w values in the
data array. They are:
nth
: nth[i] is the ith unique block to be written to since the last bulk
writeinverse_nth
: inverse_nth[i] is the order in in which the ith block of the
array was written, counting from 0 at the last bulk write.bitset
: The jth bit ofbitset[i]
is 1 when the array cell numbered
64*i + j has been written to since the last bulk write.
bitset[i]
and inverse_nth[i]
are allowed to be invalid if i
is not a
member of the set nth[0]
, nth[1]
, ... , nth[size-1]
. In other words,inverse_nth[i]
and bitset[i]
are valid if and only if inverse_nth[i] < size
and nth[inverse_nth[i]] == i
.
Rather than store three separate arrays of the same length, I chose to store one
array, is_set
, with three fields.
*/
typedef struct
int nth_;
int inverse_nth_;
uint64_t bitset_;
IsSetCell;
typedef struct
int unset_;
int size_;
IsSetCell is_set_;
IsSetArray;
typedef struct
IsSetArray * metadata_;
int payload_;
ResettableArray;
/*
To construct an array, we need a default value to return when the reading a
value that has never been written to.
*/
ResettableArray * ConstructResettableArray(int n, int unset)
ResettableArray* result =
malloc(offsetof(ResettableArray, payload_) + n * sizeof(int));
if (!result) return NULL;
n = (n + 63) / 64;
result->metadata_ =
malloc(offsetof(IsSetArray, is_set_) + n * sizeof(IsSetCell));
if (!result->metadata_)
free(result);
return NULL;
result->metadata_->size_ = 0;
result->metadata_->unset_ = unset;
return result;
void DestructResettableArray(ResettableArray * a)
if (a) free(a->metadata_);
free(a);
/*
The bulk of the algorithm is in writing and reading the metadata. AfterIsSet()
and Set()
are defined (below), reading and writing the arrays is
straightforward.
*/
bool IsSet(const IsSetArray * a, int i);
void Set(IsSetArray * a, int i);
int GetValue(const ResettableArray * a, int i)
if (!IsSet(a->metadata_, i)) return a->metadata_->unset_;
return a->payload_[i];
void SetValue(ResettableArray * a, int i, int v)
a->payload_[i] = v;
Set(a->metadata_, i);
void SetAllValues(ResettableArray * a, int v)
a->metadata_->unset_ = v;
/*
The complex part of reading and writing is the bidirectional relationship
between inverse_nth
and nth
. If they point to each other at location i
(is_set[is_set[i].inverse_nth].nth == i
) then location i contains valid data
that was written after the last bulk write, as long as is_set[i].inverse_nth <
.
size
*/
uint64_t OneBit(int i)
return UINT64_C(1) << i;
bool IsSet(const IsSetArray * a, int i)
const int cell = i/64, offset = i & 63;
const int inverse_nth = a->is_set_[cell].inverse_nth_;
return inverse_nth < a->size_ && a->is_set_[inverse_nth].nth_ == cell &&
a->is_set_[cell].bitset_ & OneBit(offset);
void Set(IsSetArray * a, int i) a->is_set_[inverse_nth].nth_ != cell)
a->is_set_[cell].inverse_nth_ = a->size_;
a->is_set_[cell].bitset_ = 0;
a->is_set_[a->size_].nth_ = cell;
++a->size_;
a->is_set_[cell].bitset_
The code never decreases size_, which I think must be wrong. Perhaps it should be zeroed in SetAllValues()?
– AnatolyVorobey
Jul 13 '18 at 6:08
add a comment |
/*
At the time I am writing this, all of the solutions on this page will double (or
more) the amount of space required to store the array. The following solution
reduces the amount of wasted space from Ω(n) to θ(n/w), where w is the number of
bits in a computer "word". On my machine, that's 64.
This prose in this answer is in C comments, so you can copy-and-paste this
answer verbatim and compile it with your C compiler.
*/
#include <stdbool.h>
#include <stddef.h>
#include <stdint.h>
#include <stdlib.h>
/*
The problem is to support reading and writing values in an array in O(1) time
along with bulk writes in which all values in the array are written at once in
O(1) time. This is possible using a technique invented by Aho, Hopcroft and
Ullman, as far as I know. I will present a version due to Gonzalo Navarro,
"Constant-Time Array Initialization in Little
Space".
The idea is to keep three metadata arrays along with the data array. We also
keep two integers: unset
, which is the last value used in a bulk write
operation, and size
, an approximation for the number of values that have been
set since the last bulk write. At all times, the number of distinct values
written since the last bulk write is between size
and w * size
.
The three metadata arrays describe information about blocks of w values in the
data array. They are:
nth
: nth[i] is the ith unique block to be written to since the last bulk
writeinverse_nth
: inverse_nth[i] is the order in in which the ith block of the
array was written, counting from 0 at the last bulk write.bitset
: The jth bit ofbitset[i]
is 1 when the array cell numbered
64*i + j has been written to since the last bulk write.
bitset[i]
and inverse_nth[i]
are allowed to be invalid if i
is not a
member of the set nth[0]
, nth[1]
, ... , nth[size-1]
. In other words,inverse_nth[i]
and bitset[i]
are valid if and only if inverse_nth[i] < size
and nth[inverse_nth[i]] == i
.
Rather than store three separate arrays of the same length, I chose to store one
array, is_set
, with three fields.
*/
typedef struct
int nth_;
int inverse_nth_;
uint64_t bitset_;
IsSetCell;
typedef struct
int unset_;
int size_;
IsSetCell is_set_;
IsSetArray;
typedef struct
IsSetArray * metadata_;
int payload_;
ResettableArray;
/*
To construct an array, we need a default value to return when the reading a
value that has never been written to.
*/
ResettableArray * ConstructResettableArray(int n, int unset)
ResettableArray* result =
malloc(offsetof(ResettableArray, payload_) + n * sizeof(int));
if (!result) return NULL;
n = (n + 63) / 64;
result->metadata_ =
malloc(offsetof(IsSetArray, is_set_) + n * sizeof(IsSetCell));
if (!result->metadata_)
free(result);
return NULL;
result->metadata_->size_ = 0;
result->metadata_->unset_ = unset;
return result;
void DestructResettableArray(ResettableArray * a)
if (a) free(a->metadata_);
free(a);
/*
The bulk of the algorithm is in writing and reading the metadata. AfterIsSet()
and Set()
are defined (below), reading and writing the arrays is
straightforward.
*/
bool IsSet(const IsSetArray * a, int i);
void Set(IsSetArray * a, int i);
int GetValue(const ResettableArray * a, int i)
if (!IsSet(a->metadata_, i)) return a->metadata_->unset_;
return a->payload_[i];
void SetValue(ResettableArray * a, int i, int v)
a->payload_[i] = v;
Set(a->metadata_, i);
void SetAllValues(ResettableArray * a, int v)
a->metadata_->unset_ = v;
/*
The complex part of reading and writing is the bidirectional relationship
between inverse_nth
and nth
. If they point to each other at location i
(is_set[is_set[i].inverse_nth].nth == i
) then location i contains valid data
that was written after the last bulk write, as long as is_set[i].inverse_nth <
.
size
*/
uint64_t OneBit(int i)
return UINT64_C(1) << i;
bool IsSet(const IsSetArray * a, int i)
const int cell = i/64, offset = i & 63;
const int inverse_nth = a->is_set_[cell].inverse_nth_;
return inverse_nth < a->size_ && a->is_set_[inverse_nth].nth_ == cell &&
a->is_set_[cell].bitset_ & OneBit(offset);
void Set(IsSetArray * a, int i) a->is_set_[inverse_nth].nth_ != cell)
a->is_set_[cell].inverse_nth_ = a->size_;
a->is_set_[cell].bitset_ = 0;
a->is_set_[a->size_].nth_ = cell;
++a->size_;
a->is_set_[cell].bitset_
The code never decreases size_, which I think must be wrong. Perhaps it should be zeroed in SetAllValues()?
– AnatolyVorobey
Jul 13 '18 at 6:08
add a comment |
/*
At the time I am writing this, all of the solutions on this page will double (or
more) the amount of space required to store the array. The following solution
reduces the amount of wasted space from Ω(n) to θ(n/w), where w is the number of
bits in a computer "word". On my machine, that's 64.
This prose in this answer is in C comments, so you can copy-and-paste this
answer verbatim and compile it with your C compiler.
*/
#include <stdbool.h>
#include <stddef.h>
#include <stdint.h>
#include <stdlib.h>
/*
The problem is to support reading and writing values in an array in O(1) time
along with bulk writes in which all values in the array are written at once in
O(1) time. This is possible using a technique invented by Aho, Hopcroft and
Ullman, as far as I know. I will present a version due to Gonzalo Navarro,
"Constant-Time Array Initialization in Little
Space".
The idea is to keep three metadata arrays along with the data array. We also
keep two integers: unset
, which is the last value used in a bulk write
operation, and size
, an approximation for the number of values that have been
set since the last bulk write. At all times, the number of distinct values
written since the last bulk write is between size
and w * size
.
The three metadata arrays describe information about blocks of w values in the
data array. They are:
nth
: nth[i] is the ith unique block to be written to since the last bulk
writeinverse_nth
: inverse_nth[i] is the order in in which the ith block of the
array was written, counting from 0 at the last bulk write.bitset
: The jth bit ofbitset[i]
is 1 when the array cell numbered
64*i + j has been written to since the last bulk write.
bitset[i]
and inverse_nth[i]
are allowed to be invalid if i
is not a
member of the set nth[0]
, nth[1]
, ... , nth[size-1]
. In other words,inverse_nth[i]
and bitset[i]
are valid if and only if inverse_nth[i] < size
and nth[inverse_nth[i]] == i
.
Rather than store three separate arrays of the same length, I chose to store one
array, is_set
, with three fields.
*/
typedef struct
int nth_;
int inverse_nth_;
uint64_t bitset_;
IsSetCell;
typedef struct
int unset_;
int size_;
IsSetCell is_set_;
IsSetArray;
typedef struct
IsSetArray * metadata_;
int payload_;
ResettableArray;
/*
To construct an array, we need a default value to return when the reading a
value that has never been written to.
*/
ResettableArray * ConstructResettableArray(int n, int unset)
ResettableArray* result =
malloc(offsetof(ResettableArray, payload_) + n * sizeof(int));
if (!result) return NULL;
n = (n + 63) / 64;
result->metadata_ =
malloc(offsetof(IsSetArray, is_set_) + n * sizeof(IsSetCell));
if (!result->metadata_)
free(result);
return NULL;
result->metadata_->size_ = 0;
result->metadata_->unset_ = unset;
return result;
void DestructResettableArray(ResettableArray * a)
if (a) free(a->metadata_);
free(a);
/*
The bulk of the algorithm is in writing and reading the metadata. AfterIsSet()
and Set()
are defined (below), reading and writing the arrays is
straightforward.
*/
bool IsSet(const IsSetArray * a, int i);
void Set(IsSetArray * a, int i);
int GetValue(const ResettableArray * a, int i)
if (!IsSet(a->metadata_, i)) return a->metadata_->unset_;
return a->payload_[i];
void SetValue(ResettableArray * a, int i, int v)
a->payload_[i] = v;
Set(a->metadata_, i);
void SetAllValues(ResettableArray * a, int v)
a->metadata_->unset_ = v;
/*
The complex part of reading and writing is the bidirectional relationship
between inverse_nth
and nth
. If they point to each other at location i
(is_set[is_set[i].inverse_nth].nth == i
) then location i contains valid data
that was written after the last bulk write, as long as is_set[i].inverse_nth <
.
size
*/
uint64_t OneBit(int i)
return UINT64_C(1) << i;
bool IsSet(const IsSetArray * a, int i)
const int cell = i/64, offset = i & 63;
const int inverse_nth = a->is_set_[cell].inverse_nth_;
return inverse_nth < a->size_ && a->is_set_[inverse_nth].nth_ == cell &&
a->is_set_[cell].bitset_ & OneBit(offset);
void Set(IsSetArray * a, int i) a->is_set_[inverse_nth].nth_ != cell)
a->is_set_[cell].inverse_nth_ = a->size_;
a->is_set_[cell].bitset_ = 0;
a->is_set_[a->size_].nth_ = cell;
++a->size_;
a->is_set_[cell].bitset_
/*
At the time I am writing this, all of the solutions on this page will double (or
more) the amount of space required to store the array. The following solution
reduces the amount of wasted space from Ω(n) to θ(n/w), where w is the number of
bits in a computer "word". On my machine, that's 64.
This prose in this answer is in C comments, so you can copy-and-paste this
answer verbatim and compile it with your C compiler.
*/
#include <stdbool.h>
#include <stddef.h>
#include <stdint.h>
#include <stdlib.h>
/*
The problem is to support reading and writing values in an array in O(1) time
along with bulk writes in which all values in the array are written at once in
O(1) time. This is possible using a technique invented by Aho, Hopcroft and
Ullman, as far as I know. I will present a version due to Gonzalo Navarro,
"Constant-Time Array Initialization in Little
Space".
The idea is to keep three metadata arrays along with the data array. We also
keep two integers: unset
, which is the last value used in a bulk write
operation, and size
, an approximation for the number of values that have been
set since the last bulk write. At all times, the number of distinct values
written since the last bulk write is between size
and w * size
.
The three metadata arrays describe information about blocks of w values in the
data array. They are:
nth
: nth[i] is the ith unique block to be written to since the last bulk
writeinverse_nth
: inverse_nth[i] is the order in in which the ith block of the
array was written, counting from 0 at the last bulk write.bitset
: The jth bit ofbitset[i]
is 1 when the array cell numbered
64*i + j has been written to since the last bulk write.
bitset[i]
and inverse_nth[i]
are allowed to be invalid if i
is not a
member of the set nth[0]
, nth[1]
, ... , nth[size-1]
. In other words,inverse_nth[i]
and bitset[i]
are valid if and only if inverse_nth[i] < size
and nth[inverse_nth[i]] == i
.
Rather than store three separate arrays of the same length, I chose to store one
array, is_set
, with three fields.
*/
typedef struct
int nth_;
int inverse_nth_;
uint64_t bitset_;
IsSetCell;
typedef struct
int unset_;
int size_;
IsSetCell is_set_;
IsSetArray;
typedef struct
IsSetArray * metadata_;
int payload_;
ResettableArray;
/*
To construct an array, we need a default value to return when the reading a
value that has never been written to.
*/
ResettableArray * ConstructResettableArray(int n, int unset)
ResettableArray* result =
malloc(offsetof(ResettableArray, payload_) + n * sizeof(int));
if (!result) return NULL;
n = (n + 63) / 64;
result->metadata_ =
malloc(offsetof(IsSetArray, is_set_) + n * sizeof(IsSetCell));
if (!result->metadata_)
free(result);
return NULL;
result->metadata_->size_ = 0;
result->metadata_->unset_ = unset;
return result;
void DestructResettableArray(ResettableArray * a)
if (a) free(a->metadata_);
free(a);
/*
The bulk of the algorithm is in writing and reading the metadata. AfterIsSet()
and Set()
are defined (below), reading and writing the arrays is
straightforward.
*/
bool IsSet(const IsSetArray * a, int i);
void Set(IsSetArray * a, int i);
int GetValue(const ResettableArray * a, int i)
if (!IsSet(a->metadata_, i)) return a->metadata_->unset_;
return a->payload_[i];
void SetValue(ResettableArray * a, int i, int v)
a->payload_[i] = v;
Set(a->metadata_, i);
void SetAllValues(ResettableArray * a, int v)
a->metadata_->unset_ = v;
/*
The complex part of reading and writing is the bidirectional relationship
between inverse_nth
and nth
. If they point to each other at location i
(is_set[is_set[i].inverse_nth].nth == i
) then location i contains valid data
that was written after the last bulk write, as long as is_set[i].inverse_nth <
.
size
*/
uint64_t OneBit(int i)
return UINT64_C(1) << i;
bool IsSet(const IsSetArray * a, int i)
const int cell = i/64, offset = i & 63;
const int inverse_nth = a->is_set_[cell].inverse_nth_;
return inverse_nth < a->size_ && a->is_set_[inverse_nth].nth_ == cell &&
a->is_set_[cell].bitset_ & OneBit(offset);
void Set(IsSetArray * a, int i) a->is_set_[inverse_nth].nth_ != cell)
a->is_set_[cell].inverse_nth_ = a->size_;
a->is_set_[cell].bitset_ = 0;
a->is_set_[a->size_].nth_ = cell;
++a->size_;
a->is_set_[cell].bitset_
answered Jan 17 '17 at 21:25
jbapplejbapple
2,6511630
2,6511630
The code never decreases size_, which I think must be wrong. Perhaps it should be zeroed in SetAllValues()?
– AnatolyVorobey
Jul 13 '18 at 6:08
add a comment |
The code never decreases size_, which I think must be wrong. Perhaps it should be zeroed in SetAllValues()?
– AnatolyVorobey
Jul 13 '18 at 6:08
The code never decreases size_, which I think must be wrong. Perhaps it should be zeroed in SetAllValues()?
– AnatolyVorobey
Jul 13 '18 at 6:08
The code never decreases size_, which I think must be wrong. Perhaps it should be zeroed in SetAllValues()?
– AnatolyVorobey
Jul 13 '18 at 6:08
add a comment |
Python example
class d:
def __init__(self, l):
self.len = l
self.g_p = [i for i in range(self.len)]
self.g_v = [0 for i in range(self.len)]
self.g_s = self.len - 1
self.g_c = 0
def getVal(self, i):
if (i < 0 or i >= self.len):
return
if (self.g_p[i] <= self.g_s):
return self.g_v[self.g_p[i]]
return self.g_c
def setVal(self, i, v):
if (i < 0 or i >= self.len):
return
if (self.g_p[i] > self.g_s):
self.g_s += 1
self.g_p[self.g_s], self.g_p[i] = self.g_p[i], self.g_p[self.g_s]
self.g_v[self.g_p[i]] = v
def setAll(self, v):
self.g_c = v
self.g_s = -1
add a comment |
Python example
class d:
def __init__(self, l):
self.len = l
self.g_p = [i for i in range(self.len)]
self.g_v = [0 for i in range(self.len)]
self.g_s = self.len - 1
self.g_c = 0
def getVal(self, i):
if (i < 0 or i >= self.len):
return
if (self.g_p[i] <= self.g_s):
return self.g_v[self.g_p[i]]
return self.g_c
def setVal(self, i, v):
if (i < 0 or i >= self.len):
return
if (self.g_p[i] > self.g_s):
self.g_s += 1
self.g_p[self.g_s], self.g_p[i] = self.g_p[i], self.g_p[self.g_s]
self.g_v[self.g_p[i]] = v
def setAll(self, v):
self.g_c = v
self.g_s = -1
add a comment |
Python example
class d:
def __init__(self, l):
self.len = l
self.g_p = [i for i in range(self.len)]
self.g_v = [0 for i in range(self.len)]
self.g_s = self.len - 1
self.g_c = 0
def getVal(self, i):
if (i < 0 or i >= self.len):
return
if (self.g_p[i] <= self.g_s):
return self.g_v[self.g_p[i]]
return self.g_c
def setVal(self, i, v):
if (i < 0 or i >= self.len):
return
if (self.g_p[i] > self.g_s):
self.g_s += 1
self.g_p[self.g_s], self.g_p[i] = self.g_p[i], self.g_p[self.g_s]
self.g_v[self.g_p[i]] = v
def setAll(self, v):
self.g_c = v
self.g_s = -1
Python example
class d:
def __init__(self, l):
self.len = l
self.g_p = [i for i in range(self.len)]
self.g_v = [0 for i in range(self.len)]
self.g_s = self.len - 1
self.g_c = 0
def getVal(self, i):
if (i < 0 or i >= self.len):
return
if (self.g_p[i] <= self.g_s):
return self.g_v[self.g_p[i]]
return self.g_c
def setVal(self, i, v):
if (i < 0 or i >= self.len):
return
if (self.g_p[i] > self.g_s):
self.g_s += 1
self.g_p[self.g_s], self.g_p[i] = self.g_p[i], self.g_p[self.g_s]
self.g_v[self.g_p[i]] = v
def setAll(self, v):
self.g_c = v
self.g_s = -1
answered Jun 11 '14 at 9:57
RoeeRoee
237
237
add a comment |
add a comment |
Regarding Timothy Jone's answer:
A problem with this approach is that eventually you'll run out of ids for timestamp, and might wrap around. If you chose a 64 bit value to store timestamps, then this gives you 18,446,744,073,709,551,616 insertions or setAlls before this happens. Depending on the expected use of the datastructure, an O(n) cleanup phase might be appropriate, or you could just throw an exception.
This is exactly the worst case scenario which make this solution O(n) too, and not O(1). This stucture, although saves a lot of potential O(n) insert operation, is still in O(n) effeciency.
See my answer for a way to maintain constant time updates in the face of overflow.
– dfeuer
Mar 6 '15 at 1:51
@dfeuer can you please specify in words how you unbox the Int and Word values? This question is being asked around the world in interviews, and the expected answer by interviewers (as descibed by Timothy) is definitly incorrect, or at least incomplete. In any case, it is obvious that the answer above should be refined. Thanks in advance.
– Emliti
Mar 6 '15 at 13:31
That sounds like a pretty good reason to delete my answer altogether.
– dfeuer
Mar 6 '15 at 13:59
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post.
– Ian
Mar 6 '15 at 14:20
@Bluehound Because I ususally don't write here I cannot leave a comment under Timothy's answer. I know I have made an important remark so please don't delete it. If pereferable move this answer under Timothy's answer so anyone can see and comment. Thank you
– Emliti
Mar 6 '15 at 19:20
|
show 22 more comments
Regarding Timothy Jone's answer:
A problem with this approach is that eventually you'll run out of ids for timestamp, and might wrap around. If you chose a 64 bit value to store timestamps, then this gives you 18,446,744,073,709,551,616 insertions or setAlls before this happens. Depending on the expected use of the datastructure, an O(n) cleanup phase might be appropriate, or you could just throw an exception.
This is exactly the worst case scenario which make this solution O(n) too, and not O(1). This stucture, although saves a lot of potential O(n) insert operation, is still in O(n) effeciency.
See my answer for a way to maintain constant time updates in the face of overflow.
– dfeuer
Mar 6 '15 at 1:51
@dfeuer can you please specify in words how you unbox the Int and Word values? This question is being asked around the world in interviews, and the expected answer by interviewers (as descibed by Timothy) is definitly incorrect, or at least incomplete. In any case, it is obvious that the answer above should be refined. Thanks in advance.
– Emliti
Mar 6 '15 at 13:31
That sounds like a pretty good reason to delete my answer altogether.
– dfeuer
Mar 6 '15 at 13:59
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post.
– Ian
Mar 6 '15 at 14:20
@Bluehound Because I ususally don't write here I cannot leave a comment under Timothy's answer. I know I have made an important remark so please don't delete it. If pereferable move this answer under Timothy's answer so anyone can see and comment. Thank you
– Emliti
Mar 6 '15 at 19:20
|
show 22 more comments
Regarding Timothy Jone's answer:
A problem with this approach is that eventually you'll run out of ids for timestamp, and might wrap around. If you chose a 64 bit value to store timestamps, then this gives you 18,446,744,073,709,551,616 insertions or setAlls before this happens. Depending on the expected use of the datastructure, an O(n) cleanup phase might be appropriate, or you could just throw an exception.
This is exactly the worst case scenario which make this solution O(n) too, and not O(1). This stucture, although saves a lot of potential O(n) insert operation, is still in O(n) effeciency.
Regarding Timothy Jone's answer:
A problem with this approach is that eventually you'll run out of ids for timestamp, and might wrap around. If you chose a 64 bit value to store timestamps, then this gives you 18,446,744,073,709,551,616 insertions or setAlls before this happens. Depending on the expected use of the datastructure, an O(n) cleanup phase might be appropriate, or you could just throw an exception.
This is exactly the worst case scenario which make this solution O(n) too, and not O(1). This stucture, although saves a lot of potential O(n) insert operation, is still in O(n) effeciency.
answered Mar 5 '15 at 13:01
EmlitiEmliti
13
13
See my answer for a way to maintain constant time updates in the face of overflow.
– dfeuer
Mar 6 '15 at 1:51
@dfeuer can you please specify in words how you unbox the Int and Word values? This question is being asked around the world in interviews, and the expected answer by interviewers (as descibed by Timothy) is definitly incorrect, or at least incomplete. In any case, it is obvious that the answer above should be refined. Thanks in advance.
– Emliti
Mar 6 '15 at 13:31
That sounds like a pretty good reason to delete my answer altogether.
– dfeuer
Mar 6 '15 at 13:59
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post.
– Ian
Mar 6 '15 at 14:20
@Bluehound Because I ususally don't write here I cannot leave a comment under Timothy's answer. I know I have made an important remark so please don't delete it. If pereferable move this answer under Timothy's answer so anyone can see and comment. Thank you
– Emliti
Mar 6 '15 at 19:20
|
show 22 more comments
See my answer for a way to maintain constant time updates in the face of overflow.
– dfeuer
Mar 6 '15 at 1:51
@dfeuer can you please specify in words how you unbox the Int and Word values? This question is being asked around the world in interviews, and the expected answer by interviewers (as descibed by Timothy) is definitly incorrect, or at least incomplete. In any case, it is obvious that the answer above should be refined. Thanks in advance.
– Emliti
Mar 6 '15 at 13:31
That sounds like a pretty good reason to delete my answer altogether.
– dfeuer
Mar 6 '15 at 13:59
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post.
– Ian
Mar 6 '15 at 14:20
@Bluehound Because I ususally don't write here I cannot leave a comment under Timothy's answer. I know I have made an important remark so please don't delete it. If pereferable move this answer under Timothy's answer so anyone can see and comment. Thank you
– Emliti
Mar 6 '15 at 19:20
See my answer for a way to maintain constant time updates in the face of overflow.
– dfeuer
Mar 6 '15 at 1:51
See my answer for a way to maintain constant time updates in the face of overflow.
– dfeuer
Mar 6 '15 at 1:51
@dfeuer can you please specify in words how you unbox the Int and Word values? This question is being asked around the world in interviews, and the expected answer by interviewers (as descibed by Timothy) is definitly incorrect, or at least incomplete. In any case, it is obvious that the answer above should be refined. Thanks in advance.
– Emliti
Mar 6 '15 at 13:31
@dfeuer can you please specify in words how you unbox the Int and Word values? This question is being asked around the world in interviews, and the expected answer by interviewers (as descibed by Timothy) is definitly incorrect, or at least incomplete. In any case, it is obvious that the answer above should be refined. Thanks in advance.
– Emliti
Mar 6 '15 at 13:31
That sounds like a pretty good reason to delete my answer altogether.
– dfeuer
Mar 6 '15 at 13:59
That sounds like a pretty good reason to delete my answer altogether.
– dfeuer
Mar 6 '15 at 13:59
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post.
– Ian
Mar 6 '15 at 14:20
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post.
– Ian
Mar 6 '15 at 14:20
@Bluehound Because I ususally don't write here I cannot leave a comment under Timothy's answer. I know I have made an important remark so please don't delete it. If pereferable move this answer under Timothy's answer so anyone can see and comment. Thank you
– Emliti
Mar 6 '15 at 19:20
@Bluehound Because I ususally don't write here I cannot leave a comment under Timothy's answer. I know I have made an important remark so please don't delete it. If pereferable move this answer under Timothy's answer so anyone can see and comment. Thank you
– Emliti
Mar 6 '15 at 19:20
|
show 22 more comments
This is my answer in Java (I'm not completly sure about the syntax).
I did the set-functions synchronized in order to avoid a situation where changeT and defChangeT are equal.
Struct ValueSt
int value;
Timestamp changeT;
Class MyDS
int default;
Timestamp defChangeT;
HashMap map;
public MyDS()
map = new HashMap<int, ValueSt>();
public synchronized void mySet(Int key, int value)
ValueSt vs = new ValueSt(value, Timestamp(System.current));
map.put(key, vs);
public synchronized void setAll(int value)
default = value;
defChangeT = Timestamp(System.current));
public int myGet(int key)
ValueSt vs = map.get(key);
if(vs != null)
if(vs.changeT > defChangeT)
return vs.value;
else
return default;
return null;
add a comment |
This is my answer in Java (I'm not completly sure about the syntax).
I did the set-functions synchronized in order to avoid a situation where changeT and defChangeT are equal.
Struct ValueSt
int value;
Timestamp changeT;
Class MyDS
int default;
Timestamp defChangeT;
HashMap map;
public MyDS()
map = new HashMap<int, ValueSt>();
public synchronized void mySet(Int key, int value)
ValueSt vs = new ValueSt(value, Timestamp(System.current));
map.put(key, vs);
public synchronized void setAll(int value)
default = value;
defChangeT = Timestamp(System.current));
public int myGet(int key)
ValueSt vs = map.get(key);
if(vs != null)
if(vs.changeT > defChangeT)
return vs.value;
else
return default;
return null;
add a comment |
This is my answer in Java (I'm not completly sure about the syntax).
I did the set-functions synchronized in order to avoid a situation where changeT and defChangeT are equal.
Struct ValueSt
int value;
Timestamp changeT;
Class MyDS
int default;
Timestamp defChangeT;
HashMap map;
public MyDS()
map = new HashMap<int, ValueSt>();
public synchronized void mySet(Int key, int value)
ValueSt vs = new ValueSt(value, Timestamp(System.current));
map.put(key, vs);
public synchronized void setAll(int value)
default = value;
defChangeT = Timestamp(System.current));
public int myGet(int key)
ValueSt vs = map.get(key);
if(vs != null)
if(vs.changeT > defChangeT)
return vs.value;
else
return default;
return null;
This is my answer in Java (I'm not completly sure about the syntax).
I did the set-functions synchronized in order to avoid a situation where changeT and defChangeT are equal.
Struct ValueSt
int value;
Timestamp changeT;
Class MyDS
int default;
Timestamp defChangeT;
HashMap map;
public MyDS()
map = new HashMap<int, ValueSt>();
public synchronized void mySet(Int key, int value)
ValueSt vs = new ValueSt(value, Timestamp(System.current));
map.put(key, vs);
public synchronized void setAll(int value)
default = value;
defChangeT = Timestamp(System.current));
public int myGet(int key)
ValueSt vs = map.get(key);
if(vs != null)
if(vs.changeT > defChangeT)
return vs.value;
else
return default;
return null;
answered Apr 10 '18 at 14:05
ChedvaChedva
93
93
add a comment |
add a comment |
The correct solution in C#:
public sealed class SpecialDictionary<T, V>
private Dictionary<T, Tuple<DateTime, V>> innerData;
private Tuple<DateTime, V> setAllValue;
private DateTime prevTime;
public SpecialDictionary()
innerData = new Dictionary<T, Tuple<DateTime, V>>();
public void Set(T key, V value) => innerData[key] = new Tuple<DateTime, V>(GetTime(), value);
public void SetAll(V value) => setAllValue = new Tuple<DateTime, V>(GetTime(), value);
public V Get(T key)
Tuple<DateTime, V> tmpValue = innerData[key];
if (setAllValue?.Item1 > tmpValue.Item1)
return setAllValue.Item2;
else
return tmpValue.Item2;
private DateTime GetTime()
if (prevTime == null)
prevTime = DateTime.Now;
else
if (prevTime == DateTime.Now)
Thread.Sleep(1);
prevTime = DateTime.Now;
return prevTime;
And the test:
static void Main(string args)
SpecialDictionary<string, int> dict = new SpecialDictionary<string, int>();
dict.Set("testA", 1);
dict.Set("testB", 2);
dict.Set("testC", 3);
Console.WriteLine(dict.Get("testC"));
dict.SetAll(4);
dict.Set("testE", 5);
Console.WriteLine(dict.Get("testC"));
Console.WriteLine(dict.Get("testE"));
Console.ReadKey();
add a comment |
The correct solution in C#:
public sealed class SpecialDictionary<T, V>
private Dictionary<T, Tuple<DateTime, V>> innerData;
private Tuple<DateTime, V> setAllValue;
private DateTime prevTime;
public SpecialDictionary()
innerData = new Dictionary<T, Tuple<DateTime, V>>();
public void Set(T key, V value) => innerData[key] = new Tuple<DateTime, V>(GetTime(), value);
public void SetAll(V value) => setAllValue = new Tuple<DateTime, V>(GetTime(), value);
public V Get(T key)
Tuple<DateTime, V> tmpValue = innerData[key];
if (setAllValue?.Item1 > tmpValue.Item1)
return setAllValue.Item2;
else
return tmpValue.Item2;
private DateTime GetTime()
if (prevTime == null)
prevTime = DateTime.Now;
else
if (prevTime == DateTime.Now)
Thread.Sleep(1);
prevTime = DateTime.Now;
return prevTime;
And the test:
static void Main(string args)
SpecialDictionary<string, int> dict = new SpecialDictionary<string, int>();
dict.Set("testA", 1);
dict.Set("testB", 2);
dict.Set("testC", 3);
Console.WriteLine(dict.Get("testC"));
dict.SetAll(4);
dict.Set("testE", 5);
Console.WriteLine(dict.Get("testC"));
Console.WriteLine(dict.Get("testE"));
Console.ReadKey();
add a comment |
The correct solution in C#:
public sealed class SpecialDictionary<T, V>
private Dictionary<T, Tuple<DateTime, V>> innerData;
private Tuple<DateTime, V> setAllValue;
private DateTime prevTime;
public SpecialDictionary()
innerData = new Dictionary<T, Tuple<DateTime, V>>();
public void Set(T key, V value) => innerData[key] = new Tuple<DateTime, V>(GetTime(), value);
public void SetAll(V value) => setAllValue = new Tuple<DateTime, V>(GetTime(), value);
public V Get(T key)
Tuple<DateTime, V> tmpValue = innerData[key];
if (setAllValue?.Item1 > tmpValue.Item1)
return setAllValue.Item2;
else
return tmpValue.Item2;
private DateTime GetTime()
if (prevTime == null)
prevTime = DateTime.Now;
else
if (prevTime == DateTime.Now)
Thread.Sleep(1);
prevTime = DateTime.Now;
return prevTime;
And the test:
static void Main(string args)
SpecialDictionary<string, int> dict = new SpecialDictionary<string, int>();
dict.Set("testA", 1);
dict.Set("testB", 2);
dict.Set("testC", 3);
Console.WriteLine(dict.Get("testC"));
dict.SetAll(4);
dict.Set("testE", 5);
Console.WriteLine(dict.Get("testC"));
Console.WriteLine(dict.Get("testE"));
Console.ReadKey();
The correct solution in C#:
public sealed class SpecialDictionary<T, V>
private Dictionary<T, Tuple<DateTime, V>> innerData;
private Tuple<DateTime, V> setAllValue;
private DateTime prevTime;
public SpecialDictionary()
innerData = new Dictionary<T, Tuple<DateTime, V>>();
public void Set(T key, V value) => innerData[key] = new Tuple<DateTime, V>(GetTime(), value);
public void SetAll(V value) => setAllValue = new Tuple<DateTime, V>(GetTime(), value);
public V Get(T key)
Tuple<DateTime, V> tmpValue = innerData[key];
if (setAllValue?.Item1 > tmpValue.Item1)
return setAllValue.Item2;
else
return tmpValue.Item2;
private DateTime GetTime()
if (prevTime == null)
prevTime = DateTime.Now;
else
if (prevTime == DateTime.Now)
Thread.Sleep(1);
prevTime = DateTime.Now;
return prevTime;
And the test:
static void Main(string args)
SpecialDictionary<string, int> dict = new SpecialDictionary<string, int>();
dict.Set("testA", 1);
dict.Set("testB", 2);
dict.Set("testC", 3);
Console.WriteLine(dict.Get("testC"));
dict.SetAll(4);
dict.Set("testE", 5);
Console.WriteLine(dict.Get("testC"));
Console.WriteLine(dict.Get("testE"));
Console.ReadKey();
answered Jun 15 '17 at 9:08
shlatchzshlatchz
7991932
7991932
add a comment |
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1
@leppie I disagree - see my answer :)
– Timothy Jones
Apr 4 '12 at 6:05
1
@TimothyJones: OK, I understand your answer, but it is 'cheating'. You are only setting 1 value, not all values.
– leppie
Apr 4 '12 at 6:08
3
@leppie The data structure would work as specified, so I'm not sure what's cheating about it ;)
– Timothy Jones
Apr 4 '12 at 7:27
Is it feasible to implement by a hash table?
– Hengameh
May 2 '15 at 8:15
@RoyiNamir: But as soon as you set a value again, you get the old values back in your code.
– leppie
Jan 10 '16 at 19:05