21. Combination Sum

Problem Statement

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: []

Solution

class Solution:
    def combinationSum(self, candidates: list[int], target: int) -> list[list[int]]:
        result = []

        def backtrack(remaining, combination, start):
            if remaining == 0:
                # Found a valid combination
                result.append(list(combination))
                return

            if remaining < 0:
                # Exceeded the target
                return

            for i in range(start, len(candidates)):
                # Add the number to the current combination
                combination.append(candidates[i])
                # Continue exploring with the same number (since we can reuse it)
                backtrack(remaining - candidates[i], combination, i)
                # Backtrack: remove the number to explore other possibilities
                combination.pop()

        backtrack(target, [], 0)
        return result

Explanation

This problem is a classic backtracking problem.

The backtrack function explores all possible combinations recursively.

  1. Base Cases:

    • If remaining (the target sum we still need to reach) is 0, it means we have found a valid combination. We add a copy of the current combination to our result.
    • If remaining is less than 0, it means the current path is invalid, so we stop exploring it.

  2. Recursive Step:

    • We iterate through the candidates starting from the start index. The start index is used to ensure that we don't generate duplicate combinations (e.g., [2,3] and [3,2] if we always started from index 0).
    • For each candidate, we add it to our current combination.
    • We then make a recursive call to backtrack, reducing the remaining sum and passing the same start index (i) because we are allowed to reuse the same element.
    • After the recursive call returns, we backtrack by removing the last added element (combination.pop()). This allows us to explore other branches of the decision tree.

The time complexity is hard to define precisely but is exponential, roughly O(N^(T/M + 1)), where N is the number of candidates, T is the target, and M is the minimum candidate value. The space complexity is O(T/M) for the recursion depth.