# Combination Sum LeetCode Solution | Easy Approach Share:

Combination Sum | Given an array of distinct integers `candidates` and a target integer `target`, 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`

Time:

Space: O(target)

### C++

`````` 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();
}
}
};
``````

### Java

`````` 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);
}
}
}

``````

### Python

`````` 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.clone())
return

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

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

``````

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