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],
[5]
]
```

Constraints:

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

Idea:

DFS

• Important Step: sort the array
• Create a DFS recursive and keep tracking target value and start index
• Exit DFS recursive:
• if (target < 0) return;
• if (target === 0) record answer and return;
• For loop (Possible solution) pay attention to:
• No duplication
• Candidates can only be used once

Solution:

``````/**
* @param {number[]} candidates
* @param {number} target
* @return {number[][]}
*/

var combinationSum2 = function(candidates, target) {
let res = [];
let cur = [];
// Important Step: sort the array
candidates.sort();
dfs(res, cur, 0, candidates, target);
return res;
}

function dfs(res, cur, start, can, target) {
// exit recursive condition
if (target < 0) return;
if (target === 0) {
res.push(cur.concat());
return;
}

for (let i = start; i < can.length; i++) {
// skip duplicate
if (i > start && can[i] === can[i - 1]) continue;
cur.push(can[i]);
// Each number in candidates may only be used once
// in the combination. start from i + 1
dfs(res, cur, i + 1, can, target - can[i]);
cur.pop();
}

}``````