# 40. Combination Sum II (Medium)

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

Given a collection of candidate numbers (`candidates`) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sums to `target`.

Each number in `candidates` may only be used once in the combination.

Note:

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

Example 1:

```Input: candidates = `[10,1,2,7,6,1,5]`, target = `8`,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]
```

Example 2:

```Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
[1,2,2],
[5]
]
```

## Solutions

``````ppublic class Solution {
public List<List<Integer>> combinationSum2(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;
}

for (int i = depth; i < nums.length; i++) {
// optimization, no need to go deeper
if (nums[i] > target) {
break;
}

// Since nums[depth] is not compared with previous one nums[depth-1].
// For combination [1,1...] will still be included and if not add following statement
// there will be a problem of duplicate answers.
if (i != depth && nums[i] == nums[i - 1]) {
continue;
}

// Since every element in nums can be chosen only once
// logic should start from i+1, not i
combine(ans, solution, nums, target - nums[i], i + 1);

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