Combination sum ii

class Solution:
    def __init__(self):
        self.path = []
        self.res = []
        self.currSum = 0
        self.used = None

    def backtrack(self, candidates, target, start):
        if self.currSum >= target:
            if self.currSum == target:
                self.res.append(self.path.copy())
            return None
        usedSet = set()
        for i in range(start, len(candidates)):
            num = candidates[i]
            if num in usedSet:
                continue
            # if i != 0 and candidates[i] == candidates[i - 1] and not self.used[i - 1]:
            #     continue
            usedSet.add(num)
            self.path.append(num)
            self.currSum += num
            # self.used[i] = True
            self.backtrack(candidates, target, i + 1)
            # self.used[i] = False
            self.currSum -= num
            self.path.pop()

    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
        candidates.sort()
        self.used = [False] * len(candidates)

        self.backtrack(candidates, target, 0)

        return self.res

Combination Sum II

Difficulty: Medium

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