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]

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 is 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 the C++ implementation of the above steps.

## C++

`// 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 << ` `")"` `;` ` ` `}` ` ` `}` `}` |

*chevron_right*

*filter_none*

**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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