# LeetCode 40. Combination Sum II

## Description

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

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

Note: The solution set must not contain duplicate combinations.

Example 1:

```Input: candidates = [10,1,2,7,6,1,5], target = 8
Output:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]
```

Example 2:

```Input: candidates = [2,5,2,1,2], target = 5
Output:
[
[1,2,2],

]
```

Constraints:

• `1 <= candidates.length <= 100`
• `1 <= candidates[i] <= 50`
• `1 <= target <= 30`

## Explanation

Combinations are subsets of the candidate array. We can use depth-first search to find all combinations of the input array. First, sort candidates array into ascending order.
Second, conduct a depth-first search to visit each combination of the input array.

## Java Solution

``````class Solution {
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
List<List<Integer>> results = new ArrayList<>();

if (candidates == null || candidates.length == 0) {
return results;
}

Arrays.sort(candidates);

List<Integer> combination = new ArrayList<>();
toFindCombinationsToTarget(candidates, results, combination, 0, target);

return results;
}

private void toFindCombinationsToTarget(int[] candidates, List<List<Integer>> results, List<Integer> combination, int startIndex, int target) {
if (target == 0) {
return;
}

for (int i = startIndex; i < candidates.length; i++) {
if (i != startIndex && candidates[i] == candidates[i - 1]) {
continue;
}

if (candidates[i] > target) {
break;
}

toFindCombinationsToTarget(candidates, results, combination, i + 1, target - candidates[i]);
combination.remove(combination.size() - 1);
}

}
}
``````

## Python Solution

``````class Solution:
def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
results = []

candidates = sorted(candidates)
self.helper(results, candidates, target, [], 0)

return results

def helper(self, results, candidates, target, combination, start):
if target == 0:
results.append(list(combination))

return

if target < 0:
return

for i in range(start, len(candidates)):
if i != start and candidates[i] == candidates[i - 1]:
continue

combination.append(candidates[i])
self.helper(results, candidates, target - candidates[i], combination, i + 1)
combination.pop()

``````
• Time Complexity: O(2^N)
• Space Complexity: O(N)