# 39. Combination Sum (Medium)

https://leetcode.com/problems/combination-sum/

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]
]
```

## Solutions

``````public class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> ans = new ArrayList<>();

if (candidates == null) {
return ans;
}

Arrays.sort(candidates);

combine(ans, new ArrayList<>(), candidates, target, 0);

return ans;
}

private void combine(List<List<Integer>> ans, List<Integer> solution, int[] nums, int target, int depth) {
if (target < 0) {
return;
}

if (target == 0) {

return;
}

// i starts from depth, it also can start from 0, but will produce duplicates
// The reason why we start from depth is that (1, 0) is equivalent with (0, 1)
for (int i = depth; i < nums.length; i++) {
// optimization, no need to go deeper
if (nums[i] > target) {
break;
}

// If the the nums contains duplicate values, skip it
// However, this problem guarantees no duplicates
if (i > depth && nums[i] == nums[i - 1]) {
continue;
}

combine(ans, solution, nums, target - nums[i], i);

solution.remove(solution.size() - 1);
}
}
}
``````