## Combination Sum II

• Time:O(n \cdot 2^n)
• Space:O(n + n \cdot 2^n) = O(n \cdot 2^n)

## C++

class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int>> ans;

sort(begin(candidates), end(candidates));
dfs(candidates, 0, target, {}, ans);
return ans;
}

private:
void dfs(const vector<int>& A, int s, int target, vector<int>&& path,
vector<vector<int>>& ans) {
if (target < 0)
return;
if (target == 0) {
ans.push_back(path);
return;
}

for (int i = s; i < A.size(); ++i) {
if (i > s && A[i] == A[i - 1])
continue;
path.push_back(A[i]);
dfs(A, i + 1, target - A[i], move(path), ans);
path.pop_back();
}
}
};


## JAVA

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

Arrays.sort(candidates);
dfs(0, candidates, target, new ArrayList<>(), ans);
return ans;
}

private void dfs(int s, int[] candidates, int target, List<Integer> path,
List<List<Integer>> ans) {
if (target < 0)
return;
if (target == 0) {
return;
}

for (int i = s; i < candidates.length; ++i) {
if (i > s && candidates[i] == candidates[i - 1])
continue;
dfs(i + 1, candidates, target - candidates[i], path, ans);
path.remove(path.size() - 1);
}
}
}


## Python

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

def dfs(s: int, target: int, path: List[int]) -> None:
if target < 0:
return
if target == 0:
ans.append(path.copy())
return

for i in range(s, len(candidates)):
if i > s and candidates[i] == candidates[i - 1]:
continue
path.append(candidates[i])
dfs(i + 1, target - candidates[i], path)
path.pop()

candidates.sort()
dfs(0, target, [])
return ans