39. Combination Sum

Problem

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.

Solution

The idea of a solution is to find a path in which each number adds to the target. All we need is to find all the distinct paths.

from typing import List

class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        path = []
        all = set()
        self.traverse(candidates, target, path, all)
        return [list(i) for i in all]


    def traverse(self, candidates: List[int], target: int, path: List[int], all: set) -> List[List[int]]:
        for n in candidates:
            new_path = [i for i in path]
            new_path.append(n)

            if target - n == 0:
                all.add(tuple(sorted(new_path)))
            elif target - n > 0:
                self.traverse(candidates, target-n, new_path, all)
            else:
                continue

A more efficient approach:

from typing import List

class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        path = []
        all = []
        self.traverse(candidates, target, path, all)
        return all

    def traverse(self, candidates: List[int], target: int, path: List[int], all: set) -> List[List[int]]:
        if target == 0:
            all.append(path)
            return
        if target < 0:
            return

        for i, n in enumerate(candidates):
            self.traverse(candidates[i:], target-n, path+[n], all)