Performance optimization nested loops - c++

I am implementing a rather complicated code and in one of the critical sections I need to basically consider all the possible strings of numbers following a certain rule. The naive implementation to explain what I do would be such a nested loop implementation:
std::array<int,3> max = { 3, 4, 6};
for(int i = 0; i <= max.at(0); ++i){
for(int j = 0; j <= max.at(1); ++j){
for(int k = 0; k <= max.at(2); ++k){
DoSomething(i, j, k);
}
}
}
Obviously I actually need more nested for and the "max" rule is more complicated but the idea is clear I think.
I implemented this idea using a recursive function approach:
std::array<int,3> max = { 3, 4, 6};
std::array<int,3> index = {0, 0, 0};
int total_depth = 3;
recursive_nested_for(0, index, max, total_depth);
where
void recursive_nested_for(int depth, std::array<int,3>& index,
std::array<int,3>& max, int total_depth)
{
if(depth != total_depth){
for(int i = 0; i <= max.at(depth); ++i){
index.at(depth) = i;
recursive_nested_for(depth+1, index, max, total_depth);
}
}
else
DoSomething(index);
}
In order to save as much as possible I declare all the variable I use global in the actual code.
Since this part of the code takes really long is it possible to do anything to speed it up?
I would also be open to write 24 nested for if necessary to avoid the overhead at least!
I thought that maybe an approach like expressions templates to actually generate at compile time these nested for could be more elegant. But is it possible?
Any suggestion would be greatly appreciated.
Thanks to all.

The recursive_nested_for() is a nice idea. It's a bit inflexible as it is currently written. However, you could use std::vector<int> for the array dimensions and indices, or make it a template to handle any size std::array<>. The compiler might be able to inline all recursive calls if it knows how deep the recursion is, and then it will probably be just as efficient as the three nested for-loops.
Another option is to use a single for loop for incrementing the indices that need incrementing:
void nested_for(std::array<int,3>& index, std::array<int,3>& max)
{
while (index.at(2) < max.at(2)) {
DoSomething(index);
// Increment indices
for (int i = 0; i < 3; ++i) {
if (++index.at(i) >= max.at(i))
index.at(i) = 0;
else
break;
}
}
}
However, you can also consider creating a linear sequence that visits all possible combinations of the iterators i, j, k and so on. For example, with array dimensions {3, 4, 6}, there are 3 * 4 * 6 = 72 possible combinations. So you can have a single counter going from 0 to 72, and then "split" that counter into the three iterator values you need, like so:
for (int c = 0; c < 72; c++) {
int k = c % 6;
int j = (c / 6) % 4;
int i = c / 6 / 4;
DoSomething(i, j, k);
}
You can generalize this to as many dimensions as you want. Of course, the more dimensions you have, the higher the cost of splitting the linear iterator. But if your array dimensions are powers of two, it might be very cheap to do so. Also, it might be that you don't need to split it at all; for example if you are calculating the sum of all elements of a multidimensional array, you don't care about the actual indices i, j, k and so on, you just want to visit all elements once. If the array is layed out linearly in memory, then you just need a linear iterator.
Of course, if you have 24 nested for loops, you'll notice that the product of all the dimension's sizes will become a very large number. If it doesn't fit in a 32 bit integer, your code is going to be very slow. If it doesn't fit into a 64 bit integer anymore, it will never finish.

Related

What is the most efficient way to repeat elements in a vector and apply a set of different functions across all elements using Eigen?

Say I have a vector containing only positive, real elements defined like this:
Eigen::VectorXd v(1.3876, 8.6983, 5.438, 3.9865, 4.5673);
I want to generate a new vector v2 that has repeated the elements in v some k times. Then I want to apply k different functions to each of the repeated elements in the vector.
For example, if v2 was v repeated 2 times and I applied floor() and ceil() as my two functions, the result based on the above vector would be a column vector with values: [1; 2; 8; 9; 5; 6; 3; 4; 4; 5]. Preserving the order of the original values is important here as well. These values are also a simplified example, in practice, I'm generating vectors v with ~100,000 or more elements and would like to make my code as vectorizable as possible.
Since I'm coming to Eigen and C++ from Matlab, the simplest approach I first took was to just convert this Nx1 vector into an Nx2 matrix, apply floor to the first column and ceil to the second column, take the transpose to get a 2xN matrix and then exploit the column-major nature of the matrix and reshape the 2xN matrix into a 2Nx1 vector, yielding the result I want. However, for large vectors, this would be very slow and inefficient.
This response by ggael effectively addresses how I could repeat the elements in the input vector by generating a sequence of indices and indexing the input vector. I could just then generate more sequences of indices to apply my functions to the relevant elements v2 and copy the result back to their respective places. However, is this really the most efficient approach? I dont fully grasp copy-on-write and move semantics, but I think the second indexing expressions would be in a sense redundant?
If that is true, then my guess is that a solution here would be some sort of nullary or unary expression where I could define an expression that accepts the vector, some index k and k expressions/functions to apply to each element and spits out the vector I'm looking for. I've read the Eigen documentation on the subject, but I'm struggling to build a functional example. Any help would be appreciated!
So, if I understand you correctly, you don't want to replicate (in terms of Eigen methods) the vector, you want to apply different methods to the same elements and store the result for each, correct?
In this case, computing it sequentially once per function is the easiest route. Most CPUs can only do one (vector) memory store per clock cycle, anyway. So for simple unary or binary operations, your gains have an upper bound.
Still, you are correct that one load is technically always better than two and it is a limitation of Eigen that there is no good way of achieving this.
Know that even if you manually write a loop that would generate multiple outputs, you should limit yourself in the number of outputs. CPUs have a limited number of line-fill buffers. IIRC Intel recommended using less than 10 "output streams" in tight loops, otherwise you could stall the CPU on those.
Another aspect is that C++'s weak aliasing restrictions make it hard for compilers to vectorize code with multiple outputs. So it might even be detrimental.
How I would structure this code
Remember that Eigen is column-major, just like Matlab. Therefore use one column per output function. Or just use separate vectors to begin with.
Eigen::VectorXd v = ...;
Eigen::MatrixX2d out(v.size(), 2);
out.col(0) = v.array().floor();
out.col(1) = v.array().ceil();
Following the KISS principle, this is good enough. You will not gain much if anything by doing something more complicated. A bit of multithreading might gain you something (less than factor 2 I would guess) because a single CPU thread is not enough to max out memory bandwidth but that's about it.
Some benchmarking
This is my baseline:
int main()
{
int rows = 100013, repetitions = 100000;
Eigen::VectorXd v = Eigen::VectorXd::Random(rows);
Eigen::MatrixX2d out(rows, 2);
for(int i = 0; i < repetitions; ++i) {
out.col(0) = v.array().floor();
out.col(1) = v.array().ceil();
}
}
Compiled with gcc-11, -O3 -mavx2 -fno-math-errno I get ca. 5.7 seconds.
Inspecting the assembler code finds good vectorization.
Plain old C++ version:
double* outfloor = out.data();
double* outceil = outfloor + out.outerStride();
const double* inarr = v.data();
for(std::ptrdiff_t j = 0; j < rows; ++j) {
const double vj = inarr[j];
outfloor[j] = std::floor(vj);
outceil[j] = std::ceil(vj);
}
40 seconds instead of 5! This version actually does not vectorize because the compiler cannot prove that the arrays don't alias each other.
Next, let's use fixed size Eigen vectors to get the compiler to generate vectorized code:
double* outfloor = out.data();
double* outceil = outfloor + out.outerStride();
const double* inarr = v.data();
std::ptrdiff_t j;
for(j = 0; j + 4 <= rows; j += 4) {
const Eigen::Vector4d vj = Eigen::Vector4d::Map(inarr + j);
const auto floorval = vj.array().floor();
const auto ceilval = vj.array().ceil();
Eigen::Vector4d::Map(outfloor + j) = floorval;
Eigen::Vector4d::Map(outceil + j) = ceilval;;
}
if(j + 2 <= rows) {
const Eigen::Vector2d vj = Eigen::Vector2d::MapAligned(inarr + j);
const auto floorval = vj.array().floor();
const auto ceilval = vj.array().ceil();
Eigen::Vector2d::Map(outfloor + j) = floorval;
Eigen::Vector2d::Map(outceil + j) = ceilval;;
j += 2;
}
if(j < rows) {
const double vj = inarr[j];
outfloor[j] = std::floor(vj);
outceil[j] = std::ceil(vj);
}
7.5 seconds. The assembler looks fine, fully vectorized. I'm not sure why performance is lower. Maybe cache line aliasing?
Last attempt: We don't try to avoid re-reading the vector but we re-read it blockwise so that it will be in cache by the time we read it a second time.
const int blocksize = 64 * 1024 / sizeof(double);
std::ptrdiff_t j;
for(j = 0; j + blocksize <= rows; j += blocksize) {
const auto& vj = v.segment(j, blocksize);
auto outj = out.middleRows(j, blocksize);
outj.col(0) = vj.array().floor();
outj.col(1) = vj.array().ceil();
}
const auto& vj = v.tail(rows - j);
auto outj = out.bottomRows(rows - j);
outj.col(0) = vj.array().floor();
outj.col(1) = vj.array().ceil();
5.4 seconds. So there is some gain here but not nearly enough to justify the added complexity.

Sort array of n elements which has k sorted sections

What is the best way to sort an section-wise sorted array as depicted in the second image?
The problem is performing a quick-sort using Message Passing Interface. The solution is performing quick-sort on array sections obtained by using MPI_Scatter() then joining the sorted
pieces using MPI_Gather().
Problem is that the array as a whole is unsorted but sections of it are.
Merging the sub-sections similarly to this solution seems like the best way of sorting the array, but considering that the sub-arrays are already within a single array other sorting algorithms may prove better.
The inputs for a sort function would be the array, it's length and the number of equally sorted sub-sections.
A signature would look something like int* sort(int* array, int length, int sections);
The sections parameter can have any value between 1 and 25. The length parameter value is greater than 0, a multiple of sections and smaller than 2^32.
This is what I am currently using:
int* merge(int* input, int length, int sections)
{
int* sub_sections_indices = new int[sections];
int* result = new int[length];
int section_size = length / sections;
for (int i = 0; i < sections; i++) //initialisation
{
sub_sections_indices[i] = 0;
}
int min, min_index, current_index;
for (int i = 0; i < length; i++) //merging
{
min_index = 0;
min = INT_MAX;
for (int j = 0; j < sections; j++)
{
if (sub_sections_indices[j] < section_size)
{
current_index = j * section_size + sub_sections_indices[j];
if (input[current_index] < min)
{
min = input[current_index];
min_index = j;
}
}
}
sub_sections_indices[min_index]++;
result[i] = min;
}
return result;
}
Optimizing for performance
I think this answer that maintains a min-heap of the smallest item of each sub-array is the best way to handle arbitrary input. However, for small values of k, think somewhere between 10 and 100, it might be faster to implement the more naive solutions given in the question you linked to; while maintaining the min-heap is only O(log n) for each step, it might have a higher overhead for small values of n than the simple linear scan from the naive solutions.
All these solutions create a copy of the input, and they maintain O(k) state.
Optimizing for space
The only way to save space I see is to sort in-place. This will be a problem for the algorithms mentioned above. An in-place algorithm will have two swap elements, but any swaps will likely destroy the property that each sub-array is sorted, unless the larger of the swapped pair is re-sorted into the sub-array it is being swapped to, which will result in an O(nĀ²) algorithm. So if you really do need to conserve memory, I think a regular in-place sorting algorithm would have to be used, which defeats your purpose.

Fastest way to find smallest missing integer from list of integers

I have a list of 100 random integers. Each random integer has a value from 0 to 99. Duplicates are allowed, so the list could be something like
56, 1, 1, 1, 1, 0, 2, 6, 99...
I need to find the smallest integer (>= 0) is that is not contained in the list.
My initial solution is this:
vector<int> integerList(100); //list of random integers
...
vector<bool> listedIntegers(101, false);
for (int theInt : integerList)
{
listedIntegers[theInt] = true;
}
int smallestInt;
for (int j = 0; j < 101; j++)
{
if (!listedIntegers[j])
{
smallestInt = j;
break;
}
}
But that requires a secondary array for book-keeping and a second (potentially full) list iteration. I need to perform this task millions of times (the actual application is in a greedy graph coloring algorithm, where I need to find the smallest unused color value with a vertex adjacency list), so I'm wondering if there's a clever way to get the same result without so much overhead?
It's been a year, but ...
One idea that comes to mind is to keep track of the interval(s) of unused values as you iterate the list. To allow efficient lookup, you could keep intervals as tuples in a binary search tree, for example.
So, using your sample data:
56, 1, 1, 1, 1, 0, 2, 6, 99...
You would initially have the unused interval [0..99], and then, as each input value is processed:
56: [0..55][57..99]
1: [0..0][2..55][57..99]
1: no change
1: no change
1: no change
0: [2..55][57..99]
2: [3..55][57..99]
6: [3..5][7..55][57..99]
99: [3..5][7..55][57..98]
Result (lowest value in lowest remaining interval): 3
I believe there is no faster way to do it. What you can do in your case is to reuse vector<bool>, you need to have just one such vector per thread.
Though the better approach might be to reconsider the whole algorithm to eliminate this step at all. Maybe you can update least unused color on every step of the algorithm?
Since you have to scan the whole list no matter what, the algorithm you have is already pretty good. The only improvement I can suggest without measuring (that will surely speed things up) is to get rid of your vector<bool>, and replace it with a stack-allocated array of 4 32-bit integers or 2 64-bit integers.
Then you won't have to pay the cost of allocating an array on the heap every time, and you can get the first unused number (the position of the first 0 bit) much faster. To find the word that contains the first 0 bit, you only need to find the first one that isn't the maximum value, and there are bit twiddling hacks you can use to get the first 0 bit in that word very quickly.
You program is already very efficient, in O(n). Only marginal gain can be found.
One possibility is to divide the number of possible values in blocks of size block, and to register
not in an array of bool but in an array of int, in this case memorizing the value modulo block.
In practice, we replace a loop of size N by a loop of size N/block plus a loop of size block.
Theoretically, we could select block = sqrt(N) = 12 in order to minimize the quantity N/block + block.
In the program hereafter, block of size 8 are selected, assuming that dividing integers by 8 and calculating values modulo 8 should be fast.
However, it is clear that a gain, if any, can be obtained only for a minimum value rather large!
constexpr int N = 100;
int find_min1 (const std::vector<int> &IntegerList) {
constexpr int Size = 13; //N / block
constexpr int block = 8;
constexpr int Vmax = 255; // 2^block - 1
int listedBlocks[Size] = {0};
for (int theInt : IntegerList) {
listedBlocks[theInt / block] |= 1 << (theInt % block);
}
for (int j = 0; j < Size; j++) {
if (listedBlocks[j] == Vmax) continue;
int &k = listedBlocks[j];
for (int b = 0; b < block; b++) {
if ((k%2) == 0) return block * j + b;
k /= 2;
}
}
return -1;
}
Potentially you can reduce the last step to O(1) by using some bit manipulation, in your case __int128, set the corresponding bits in loop one and call something like __builtin_clz or use the appropriate bit hack
The best solution I could find for finding smallest integer from a set is https://codereview.stackexchange.com/a/179042/31480
Here are c++ version.
int solution(std::vector<int>& A)
{
for (std::vector<int>::size_type i = 0; i != A.size(); i++)
{
while (0 < A[i] && A[i] - 1 < A.size()
&& A[i] != i + 1
&& A[i] != A[A[i] - 1])
{
int j = A[i] - 1;
auto tmp = A[i];
A[i] = A[j];
A[j] = tmp;
}
}
for (std::vector<int>::size_type i = 0; i != A.size(); i++)
{
if (A[i] != i+1)
{
return i + 1;
}
}
return A.size() + 1;
}

Optimizing this code block

for (int i = 0; i < 5000; i++)
for (int j = 0; j < 5000; j++)
{
for (int ii = 0; ii < 20; ii++)
for (int jj = 0; jj < 20; jj++)
{
int num = matBigger[i+ii][j+jj];
// Extract range from this.
int low = num & 0xff;
int high = num >> 8;
if (low < matSmaller[ii][jj] && matSmaller[ii][jj] > high)
// match found
}
}
The machine is x86_64, 32kb L1 cahce, 256 Kb L2 cache.
Any pointers on how can I possibly optimize this code?
EDIT Some background to the original problem : Fastest way to Find a m x n submatrix in M X N matrix
First thing I'd try is to move the ii and jj loops outside the i and j loops. That way you're using the same elements of matSmaller for 25 million iterations of the i and j loops, meaning that you (or the compiler if you're lucky) can hoist the access to them outside those loops:
for (int ii = 0; ii < 20; ii++)
for (int jj = 0; jj < 20; jj++)
int smaller = matSmaller[ii][jj];
for (int i = 0; i < 5000; i++)
for (int j = 0; j < 5000; j++) {
int num = matBigger[i+ii][j+jj];
int low = num & 0xff;
if (low < smaller && smaller > (num >> 8)) {
// match found
}
}
This might be faster (thanks to less access to the matSmaller array), or it might be slower (because I've changed the pattern of access to the matBigger array, and it's possible that I've made it less cache-friendly). A similar alternative would be to move the ii loop outside i and j and hoist matSmaller[ii], but leave the jj loop inside. The rule of thumb is that it's more cache-friendly to increment the last index of a multi-dimensional array in your inner loops, than earlier indexes. So we're "happier" to modify jj and j than we are to modify ii and i.
Second thing I'd try - what's the type of matBigger? Looks like the values in it are only 16 bits, so try it both as int and as (u)int16_t. The former might be faster because aligned int access is fast. The latter might be faster because more of the array fits in cache at any one time.
There are some higher-level things you could consider with some early analysis of smaller: for example if it's 0 then you needn't examine matBigger for that value of ii and jj, because num & 0xff < 0 is always false.
To do better than "guess things and see whether they're faster or not" you need to know for starters which line is hottest, which means you need a profiler.
Some basic advice:
Profile it, so you can learn where the hot-spots are.
Think about cache locality, and the addresses resulting from your loop order.
Use more const in the innermost scope, to hint more to the compiler.
Try breaking it up so you don't compute high if the low test is failing.
Try maintaining the offset into matBigger and matSmaller explicitly, to the innermost stepping into a simple increment.
Best thing ist to understand what the code is supposed to do, then check whether another algorithm exists for this problem.
Apart from that:
if you are just interested if a matching entry exists, make sure to break out of all 3 loops at the position of // match found.
make sure the data is stored in an optimal way. It all depends on your problem, but i.e. it could be more efficient to have just one array of size 5000*5000*20 and overload operator()(int,int,int) for accessing elements.
What are matSmaller and matBigger?
Try changing them to matBigger[i+ii * COL_COUNT + j+jj]
I agree with Steve about rearranging your loops to have the higher count as the inner loop. Since your code is only doing loads and compares, I believe a significant portion of the time is used for pointer arithmetic. Try an experiment to change Steve's answer into this:
for (int ii = 0; ii < 20; ii++)
{
for (int jj = 0; jj < 20; jj++)
{
int smaller = matSmaller[ii][jj];
for (int i = 0; i < 5000; i++)
{
int *pI = &matBigger[i+ii][jj];
for (int j = 0; j < 5000; j++)
{
int num = *pI++;
int low = num & 0xff;
if (low < smaller && smaller > (num >> 8)) {
// match found
} // for j
} // for i
} // for jj
} // for ii
Even in 64-bit mode, the C compiler doesn't necessarily do a great job of keeping everything in register. By changing the array access to be a simple pointer increment, you'll make the compiler's job easier to produce efficient code.
Edit: I just noticed #unwind suggested basically the same thing. Another issue to consider is the statistics of your comparison. Is the low or high comparison more probable? Arrange the conditional statement so that the less probable test is first.
Looks like there is a lot of repetition here. One optimization is to reduce the amount of duplicate effort. Using pen and paper, I'm showing the matBigger "i" index iterating as:
[0 + 0], [0 + 1], [0 + 2], ..., [0 + 19],
[1 + 0], [1 + 1], ..., [1 + 18], [1 + 19]
[2 + 0], ..., [2 + 17], [2 + 18], [2 + 19]
As you can see there are locations that are accessed many times.
Also, multiplying the iteration counts indicate that the inner content is accessed: 20 * 20 * 5000 * 5000, or 10000000000 (10E+9) times. That's a lot!
So rather than trying to speed up the execution of 10E9 instructions (such as execution (pipeline) cache or data cache optimization), try reducing the number of iterations.
The code is searcing the matrix for a number that is within a range: larger than a minimal value and less than the maximum range value.
Based on this, try a different approach:
Find and remember all coordinates where the search value is greater
than the low value. Let us call these anchor points.
For each anchor point, find the coordinates of the first value after
the anchor point that is outside the range.
The objective is to reduce the number of duplicate accesses. Anchor points allow for a one pass scan and allow other decisions such as finding a range or determining an MxN matrix that contains the anchor value.
Another idea is to create new data structures containing the matBigger and matSmaller that are more optimized for searching.
For example, create a {value, coordinate list} entry for each unique value in matSmaller:
Value coordinate list
26 -> (2,3), (6,5), ..., (1007, 75)
31 -> (4,7), (2634, 5), ...
Now you can use this data structure to find values in matSmaller and immediately know their locations. So you could search matBigger for each unique value in this data structure. This again reduces the number of access to the matrices.

Finding repeating signed integers with O(n) in time and O(1) in space

(This is a generalization of: Finding duplicates in O(n) time and O(1) space)
Problem: Write a C++ or C function with time and space complexities of O(n) and O(1) respectively that finds the repeating integers in a given array without altering it.
Example: Given {1, 0, -2, 4, 4, 1, 3, 1, -2} function must print 1, -2, and 4 once (in any order).
EDIT: The following solution requires a duo-bit (to represent 0, 1, and 2) for each integer in the range of the minimum to the maximum of the array. The number of necessary bytes (regardless of array size) never exceeds (INT_MAX ā€“ INT_MIN)/4 + 1.
#include <stdio.h>
void set_min_max(int a[], long long unsigned size,\
int* min_addr, int* max_addr)
{
long long unsigned i;
if(!size) return;
*min_addr = *max_addr = a[0];
for(i = 1; i < size; ++i)
{
if(a[i] < *min_addr) *min_addr = a[i];
if(a[i] > *max_addr) *max_addr = a[i];
}
}
void print_repeats(int a[], long long unsigned size)
{
long long unsigned i;
int min, max = min;
long long diff, q, r;
char* duos;
set_min_max(a, size, &min, &max);
diff = (long long)max - (long long)min;
duos = calloc(diff / 4 + 1, 1);
for(i = 0; i < size; ++i)
{
diff = (long long)a[i] - (long long)min; /* index of duo-bit
corresponding to a[i]
in sequence of duo-bits */
q = diff / 4; /* index of byte containing duo-bit in "duos" */
r = diff % 4; /* offset of duo-bit */
switch( (duos[q] >> (6 - 2*r )) & 3 )
{
case 0: duos[q] += (1 << (6 - 2*r));
break;
case 1: duos[q] += (1 << (6 - 2*r));
printf("%d ", a[i]);
}
}
putchar('\n');
free(duos);
}
void main()
{
int a[] = {1, 0, -2, 4, 4, 1, 3, 1, -2};
print_repeats(a, sizeof(a)/sizeof(int));
}
The definition of big-O notation is that its argument is a function (f(x)) that, as the variable in the function (x) tends to infinity, there exists a constant K such that the objective cost function will be smaller than Kf(x). Typically f is chosen to be the smallest such simple function such that the condition is satisfied. (It's pretty obvious how to lift the above to multiple variables.)
This matters because that K ā€” which you aren't required to specify ā€” allows a whole multitude of complex behavior to be hidden out of sight. For example, if the core of the algorithm is O(n2), it allows all sorts of other O(1), O(logn), O(n), O(nlogn), O(n3/2), etc. supporting bits to be hidden, even if for realistic input data those parts are what actually dominate. That's right, it can be completely misleading! (Some of the fancier bignum algorithms have this property for real. Lying with mathematics is a wonderful thing.)
So where is this going? Well, you can assume that int is a fixed size easily enough (e.g., 32-bit) and use that information to skip a lot of trouble and allocate fixed size arrays of flag bits to hold all the information that you really need. Indeed, by using two bits per potential value (one bit to say whether you've seen the value at all, another to say whether you've printed it) then you can handle the code with fixed chunk of memory of 1GB in size. That will then give you enough flag information to cope with as many 32-bit integers as you might ever wish to handle. (Heck that's even practical on 64-bit machines.) Yes, it's going to take some time to set that memory block up, but it's constant so it's formally O(1) and so drops out of the analysis. Given that, you then have constant (but whopping) memory consumption and linear time (you've got to look at each value to see whether it's new, seen once, etc.) which is exactly what was asked for.
It's a dirty trick though. You could also try scanning the input list to work out the range allowing less memory to be used in the normal case; again, that adds only linear time and you can strictly bound the memory required as above so that's constant. Yet more trickiness, but formally legal.
[EDIT] Sample C code (this is not C++, but I'm not good at C++; the main difference would be in how the flag arrays are allocated and managed):
#include <stdio.h>
#include <stdlib.h>
// Bit fiddling magic
int is(int *ary, unsigned int value) {
return ary[value>>5] & (1<<(value&31));
}
void set(int *ary, unsigned int value) {
ary[value>>5] |= 1<<(value&31);
}
// Main loop
void print_repeats(int a[], unsigned size) {
int *seen, *done;
unsigned i;
seen = calloc(134217728, sizeof(int));
done = calloc(134217728, sizeof(int));
for (i=0; i<size; i++) {
if (is(done, (unsigned) a[i]))
continue;
if (is(seen, (unsigned) a[i])) {
set(done, (unsigned) a[i]);
printf("%d ", a[i]);
} else
set(seen, (unsigned) a[i]);
}
printf("\n");
free(done);
free(seen);
}
void main() {
int a[] = {1,0,-2,4,4,1,3,1,-2};
print_repeats(a,sizeof(a)/sizeof(int));
}
Since you have an array of integers you can use the straightforward solution with sorting the array (you didn't say it can't be modified) and printing duplicates. Integer arrays can be sorted with O(n) and O(1) time and space complexities using Radix sort. Although, in general it might require O(n) space, the in-place binary MSD radix sort can be trivially implemented using O(1) space (look here for more details).
The O(1) space constraint is intractable.
The very fact of printing the array itself requires O(N) storage, by definition.
Now, feeling generous, I'll give you that you can have O(1) storage for a buffer within your program and consider that the space taken outside the program is of no concern to you, and thus that the output is not an issue...
Still, the O(1) space constraint feels intractable, because of the immutability constraint on the input array. It might not be, but it feels so.
And your solution overflows, because you try to memorize an O(N) information in a finite datatype.
There is a tricky problem with definitions here. What does O(n) mean?
Konstantin's answer claims that the radix sort time complexity is O(n). In fact it is O(n log M), where the base of the logarithm is the radix chosen, and M is the range of values that the array elements can have. So, for instance, a binary radix sort of 32-bit integers will have log M = 32.
So this is still, in a sense, O(n), because log M is a constant independent of n. But if we allow this, then there is a much simpler solution: for each integer in the range (all 4294967296 of them), go through the array to see if it occurs more than once. This is also, in a sense, O(n), because 4294967296 is also a constant independent of n.
I don't think my simple solution would count as an answer. But if not, then we shouldn't allow the radix sort, either.
I doubt this is possible. Assuming there is a solution, let's see how it works. I'll try to be as general as I can and show that it can't work... So, how does it work?
Without losing generality we could say we process the array k times, where k is fixed. The solution should also work when there are m duplicates, with m >> k. Thus, in at least one of the passes, we should be able to output x duplicates, where x grows when m grows. To do so, some useful information has been computed in a previous pass and stored in the O(1) storage. (The array itself can't be used, this would give O(n) storage.)
The problem: we have O(1) of information, when we walk over the array we have to identify x numbers(to output them). We need a O(1) storage than can tell us in O(1) time, if an element is in it. Or said in a different way, we need a data structure to store n booleans (of wich x are true) that uses O(1) space, and takes O(1) time to query.
Does this data structure exists? If not, then we can't find all duplicates in an array with O(n) time and O(1) space (or there is some fancy algorithm that works in a completely different manner???).
I really don't see how you can have only O(1) space and not modify the initial array. My guess is that you need an additional data structure. For example, what is the range of the integers? If it's 0..N like in the other question you linked, you can have an additinal count array of size N. Then in O(N) traverse the original array and increment the counter at the position of the current element. Then traverse the other array and print the numbers with count >= 2. Something like:
int* counts = new int[N];
for(int i = 0; i < N; i++) {
counts[input[i]]++;
}
for(int i = 0; i < N; i++) {
if(counts[i] >= 2) cout << i << " ";
}
delete [] counts;
Say you can use the fact you are not using all the space you have. You only need one more bit per possible value and you have lots of unused bit in your 32-bit int values.
This has serious limitations, but works in this case. Numbers have to be between -n/2 and n/2 and if they repeat m times, they will be printed m/2 times.
void print_repeats(long a[], unsigned size) {
long i, val, pos, topbit = 1 << 31, mask = ~topbit;
for (i = 0; i < size; i++)
a[i] &= mask;
for (i = 0; i < size; i++) {
val = a[i] & mask;
if (val <= mask/2) {
pos = val;
} else {
val += topbit;
pos = size + val;
}
if (a[pos] < 0) {
printf("%d\n", val);
a[pos] &= mask;
} else {
a[pos] |= topbit;
}
}
}
void main() {
long a[] = {1, 0, -2, 4, 4, 1, 3, 1, -2};
print_repeats(a, sizeof (a) / sizeof (long));
}
prints
4
1
-2