Combinational Sum

Given an array of positive integers arr[] and a sum x, find all unique combinations in arr[] where the sum is equal to x. The same repeated number may be chosen from arr[] unlimited number of times. Elements in a combination (a1, a2, …, ak) must be printed in non-descending order. (ie, a1 <= a2 <= … <= ak).
The combinations themselves must be sorted in ascending order, i.e., the combination with smallest first element should be printed first. If there is no combination possible the print "Empty" (without quotes).

Examples:

```Input : arr[] = 2, 4, 6, 8
x = 8
Output : [2, 2, 2, 2]
[2, 2, 4]
[2, 6]
[4, 4]
[8]
```

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Since the problem is to get all the possible results, not the best or the number of result, thus we don’t need to consider DP(dynamic programming), recursion is needed to handle it.

We should use the following algorithm.

```1. Sort the array(non-decreasing).
2. First remove all the duplicates from array.
3. Then use recursion and backtracking to solve
the problem.
(A) If at any time sub-problem sum == 0 then
add that array to the result (vector of
vectors).
(B) Else if sum if negative then ignore that
sub-problem.
(C) Else insert the present array in that
index to the current vector and call
the function with sum = sum-ar[index] and
index = index, then pop that element from
current index (backtrack) and call the
function with sum = sum and index = index+1 ```

Below is C++ implementation of above steps.

```// C++ program to find all combinations that
// sum to a given value
#include <bits/stdc++.h>
using namespace std;

// Print all members of ar[] that have given
void findNumbers(vector<int>& ar, int sum,
vector<vector<int> >& res,
vector<int>& r, int i)
{
// If  current sum becomes negative
if (sum < 0)
return;

// if we get exact answer
if (sum == 0)
{
res.push_back(r);
return;
}

// Recur for all remaining elements that
// have value smaller than sum.
while (i < ar.size() && sum - ar[i] >= 0)
{

// Till every element in the array starting
// from i which can contribute to the sum
r.push_back(ar[i]); // add them to list

// recur for next numbers
findNumbers(ar, sum - ar[i], res, r, i);
i++;

// remove number from list (backtracking)
r.pop_back();
}
}

// Returns all combinations of ar[] that have given
// sum.
vector<vector<int> > combinationSum(vector<int>& ar,
int sum)
{
// sort input array
sort(ar.begin(), ar.end());

// remove duplicates
ar.erase(unique(ar.begin(), ar.end()), ar.end());

vector<int> r;
vector<vector<int> > res;
findNumbers(ar, sum, res, r, 0);

return res;
}

// Driver code
int main()
{
vector<int> ar;
ar.push_back(2);
ar.push_back(4);
ar.push_back(6);
ar.push_back(8);
int n = ar.size();

int sum = 8; // set value of sum
vector<vector<int> > res = combinationSum(ar, sum);

// If result is empty, then
if (res.size() == 0)
{
cout << "Emptyn";
return 0;
}

// Print all combinations stored in res.
for (int i = 0; i < res.size(); i++)
{
if (res[i].size() > 0)
{
cout << " ( ";
for (int j = 0; j < res[i].size(); j++)
cout << res[i][j] << " ";
cout << ")";
}
}
}
```

Output:

```( 2 2 2 2 ) ( 2 2 4 ) ( 2 6 ) ( 4 4 ) ( 8 )
```

This article is contributed by Aditya Nihal Kumar Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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