Introduction

In this blog post, we will explore how to solve the LeetCode problem "216" 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

Find all valid combinations of k numbers that sum up to n such that the following conditions are true: Only numbers 1 through 9 are used. Each number is used at most once. Return a list of all possible valid combinations. The list must not contain the same combination twice, and the combinations may be returned in any order. Example 1: Input: k = 3, n = 7 Output: [[1,2,4]] Explanation: 1 + 2 + 4 = 7 There are no other valid combinations. Example 2: Input: k = 3, n = 9 Output: [[1,2,6],[1,3,5],[2,3,4]] Explanation: 1 + 2 + 6 = 9 1 + 3 + 5 = 9 2 + 3 + 4 = 9 There are no other valid combinations. Example 3: Input: k = 4, n = 1 Output: [] Explanation: There are no valid combinations. Using 4 different numbers in the range [1,9], the smallest sum we can get is 1+2+3+4 = 10 and since 10 > 1, there are no valid combination. Constraints: 2 <= k <= 9 1 <= n <= 60

Explanation

• Backtracking: Use a recursive backtracking approach to explore all possible combinations of numbers from 1 to 9.
  • Pruning: Optimize the search by pruning branches that cannot lead to a valid solution. This includes checking if the current sum exceeds n or if the number of elements in the combination exceeds k.
  • Avoiding Duplicates: Start the next number in the combination from the last added number + 1, to avoid duplicates and maintain the ordering within each combination.

• Runtime Complexity: O(C(9, k)), where C(n, k) is the binomial coefficient "n choose k". The complexity is exponential due to the nature of the problem. Storage Complexity: O(k), mainly due to the depth of the recursion stack and storing the current combination. In the worst case where all combinations are valid the storage complexity could also be O(number of combinations) to store the result.

Code

    def combinationSum3(k: int, n: int) -> list[list[int]]:
    """
    Finds all valid combinations of k numbers that sum up to n such that:
    - Only numbers 1 through 9 are used.
    - Each number is used at most once.

    Args:
        k: The number of elements in each combination.
        n: The target sum.

    Returns:
        A list of all possible valid combinations.
    """

    result = []

    def backtrack(combination, remaining_sum, start_num):
        """
        Recursive backtracking helper function.

        Args:
            combination: The current combination being built.
            remaining_sum: The remaining sum needed to reach n.
            start_num: The starting number for the next element in the combination.
        """

        if len(combination) == k:
            if remaining_sum == 0:
                result.append(combination.copy())  # Add a copy to avoid modification
            return

        if len(combination) > k or remaining_sum < 0:
            return  # Prune invalid branches

        for i in range(start_num, 10):
            combination.append(i)
            backtrack(combination, remaining_sum - i, i + 1)
            combination.pop()  # Backtrack: Remove the last added number

    backtrack([], n, 1)
    return result