Combination sum II

We will run a DFS for the decision tree and use backtracking if we failed to find a result. As we are bit allowed to use the same items twice, we will keep track of the previous item, and if it is already present in the combination, we will skip it.

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

        prev = -1
        def dfs(pos, total, current):
            if total == 0:
                res.append(current.copy())
            if total < 0:
                return

            prev = -1
            for i in range(pos, len(candidates)):
                if candidates[i] == prev:
                    continue
                prev = candidates[i]
                current.append(candidates[i])
                dfs(i+1, total-candidates[i], current)
                current.pop()

        dfs(0, target, [])
        return res

Time Complexity: O(2^n)
Space Complexity: O(1)