39. Combination Sum

MediumLeetCode

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

How to solve the problem

• Backtracking (DFS)

Code

class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        path = []
        result = []
        candidates.sort() # Timsort

        def backtracking(startIndex):
            if sum(path) == target:
                result.append(path[:])
                return # Prune
            if sum(path) > target: # Prune
                return 
            for i in range(startIndex, len(candidates)):
                path.append(candidates[i])
                backtracking(i) # Recurse with the same index (since the same number can be used unlimited times)
                path.pop()
        backtracking(0)
        return result

Complexity

• Time complexity: O(n^(target / min))
• Space complexity: O(target / min)