# Combination Sum

## Description

Given a **set** of candidate numbers (`candidates`) **(without duplicates)** and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sums to `target`.

The **same** repeated number may be chosen from `candidates` unlimited number of times.

**Note:**

* All numbers (including `target`) will be positive integers.
* The solution set must not contain duplicate combinations.

**Example 1:**

```
Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]
```

**Example 2:**

```
Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]
```

## Solution

```cpp
class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<int> nums;
        vector<vector<int>> ans;
        dfs(candidates, 0, target, nums, ans);
        return ans;
    }
    
    void dfs(vector<int> &candidates, int k, int target, vector<int> &nums, vector<vector<int>> &ans){
        if(target == 0){
            ans.push_back(nums);
            return;
        }
        if(k == candidates.size() || target < 0) return;
        int size = nums.size();
        int num = candidates[k];
        // use 0, 1, 2, ... candidates[i]
        for(int j = 0; j * num <= target; ++j){
            dfs(candidates, k + 1, target - j * num, nums, ans);
            nums.push_back(num);
        }
        nums.resize(size);
    }
};
```
