39. Combination Sum

Description

Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.

The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.

The test cases are generated such that the number of unique combinations that sum up to target is less than 150 combinations for the given input.

Example 1:

• Input: candidates = [2,3,6,7], target = 7
• Output: [[2,2,3],[7]]
• Explanation:
  2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
  7 is a candidate, and 7 = 7.
  These are the only two combinations.

Example 2:

• Input: candidates = [2,3,5], target = 8
• Output: [[2,2,2,2],[2,3,3],[3,5]]

Example 3:

• Input: candidates = [2], target = 1
• Output: []

Constraints:

• 1 <= candidates.length <= 30
• 2 <= candidates[i] <= 40
• All elements of candidates are distinct.
• 1 <= target <= 40

Submitted Code

class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        def dfs(start, total, comb):
            if total == target:
                res.append(comb.copy())
                return

            for i in range(start, len(candidates)):
                if total + candidates[i] > target:      # 그 뒤 조합 스킵
                    break
                comb.append(candidates[i])              
                dfs(i, total + candidates[i], comb)     # 현재 숫자 선택 후 다시 재귀호출
                comb.pop()                              # 원래 조합으로 복귀

        candidates.sort()
        res = []
        dfs(0, 0, [])

        return res

Runtime: 3 ms | Beats 95.20%
Memory: 19.74 MB | Beats 25.28%

현재 상태를 저장했다가 다시 되돌아오는 과정이 필요하기 때문에 백트래킹이 적합한 문제다.

Other Solutions

1st

class Solution(object):
    def combinationSum(self, candidates, target):
        ret = []
        self.dfs(candidates, target, [], ret)
        return ret
    
    def dfs(self, nums, target, path, ret):
        if target < 0:
            return 
        if target == 0:
            ret.append(path)
            return 
        for i in range(len(nums)):
            self.dfs(nums[i:], target-nums[i], path+[nums[i]], ret)

time complexity: 𝑂(𝑛^{(𝑡/𝑚𝑖𝑛)}) ← candidates’s min
space complexity: 𝑂(𝑡/𝑚𝑖𝑛)

total을 따로 저장하지 않고 target에서 값을 깎는 방법으로 풀 수도 있다.