# Combination Sum LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Combination Sum in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemCombination Sum– LeetCode Problem

`Combination Sum– LeetCode Problem`

### Problem:

Given an array of distinct integers `candidates` and a target integer, return a list of all unique combinations of `candidates` where the chosen numbers sum to `target`. You may return the combinations in any order.

The same number may be chosen from `candidates` an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.

It is guaranteed that the number of unique combinations that sum up to `target` is less than `150` combinations for the given input.

Example 1:

```Input: candidates = [2,3,6,7], target = 7
Output: [[2,2,3],]
Explanation:
2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
7 is a candidate, and 7 = 7.
These are the only two combinations.
```

Example 2:

```Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2],[2,3,3],[3,5]]
```

Example 3:

```Input: candidates = , target = 1
Output: []
```

Constraints:

• `1 <= candidates.length <= 30`
• `1 <= candidates[i] <= 200`
• All elements of `candidates` are distinct.
• `1 <= target <= 500`
`Combination Sum– LeetCode Solutions`
```class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<vector<int>> ans;

sort(begin(candidates), end(candidates));
dfs(candidates, 0, target, {}, ans);

return ans;
}

private:
void dfs(const vector<int>& A, int s, int target, vector<int>&& path,
vector<vector<int>>& ans) {
if (target < 0)
return;
if (target == 0) {
ans.push_back(path);
return;
}

for (int i = s; i < A.size(); ++i) {
path.push_back(A[i]);
dfs(A, i, target - A[i], move(path), ans);
path.pop_back();
}
}
};```
```class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> ans = new ArrayList<>();

Arrays.sort(candidates);
dfs(0, candidates, target, new ArrayList<>(), ans);

return ans;
}

private void dfs(int s, int[] candidates, int target, List<Integer> path,
List<List<Integer>> ans) {
if (target < 0)
return;
if (target == 0) {
return;
}

for (int i = s; i < candidates.length; ++i) {
dfs(i, candidates, target - candidates[i], path, ans);
path.remove(path.size() - 1);
}
}
}```
```class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
ans = []

def dfs(s: int, target: int, path: List[int]) -> None:
if target < 0:
return
if target == 0:
ans.append(path)
return

for i in range(s, len(candidates)):
dfs(i, target - candidates[i], path + [candidates[i]])

candidates.sort()
dfs(0, target, [])

return ans``` #### Ads Blocker Detected!!!

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