Introduction

In this blog post, we will explore how to solve the LeetCode problem "39" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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

Explanation

• Recursive Backtracking: The core idea is to explore possible combinations recursively. At each step, we choose a candidate number, and recursively try to find combinations for the remaining target value.
  • Allowing Repetition: Since we can use the same number multiple times, we don't move to the next candidate after using one; we stay at the same candidate and try it again.
  • Pruning: If at any point the current sum exceeds the target, we backtrack to avoid exploring unnecessary branches.

• Time Complexity: O(N^(T/M)), where N is the number of candidates, T is the target, and M is the minimum value in candidates. In the worst case, we might explore all possible combinations, but the constraint that the number of valid combinations is less than 150 helps bound the runtime. Space Complexity: O(T), due to the depth of the recursion stack and the space used to store the current combination.

Code

    def combinationSum(candidates, target):
    result = []

    def backtrack(combination, remaining, start):
        if remaining == 0:
            result.append(combination.copy())
            return
        elif remaining < 0:
            return

        for i in range(start, len(candidates)):
            combination.append(candidates[i])
            backtrack(combination, remaining - candidates[i], i)
            combination.pop()  # Backtrack: remove the last added element

    backtrack([], target, 0)
    return result