Introduction

In this blog post, we will explore how to solve the LeetCode problem "77" 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 two integers n and k, return all possible combinations of k numbers chosen from the range [1, n]. You may return the answer in any order. Example 1: Input: n = 4, k = 2 Output: [[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]] Explanation: There are 4 choose 2 = 6 total combinations. Note that combinations are unordered, i.e., [1,2] and [2,1] are considered to be the same combination. Example 2: Input: n = 1, k = 1 Output: [[1]] Explanation: There is 1 choose 1 = 1 total combination. Constraints: 1 <= n <= 20 1 <= k <= n

Explanation

Here's a breakdown of the solution and the code:

• High-level Approach:

  • Use a recursive backtracking approach to explore all possible combinations.
  • Maintain a current combination list and iteratively add numbers from the range [1, n].
  • Ensure that the numbers added to the combination are in increasing order to avoid duplicates (e.g., [1, 2] and [2, 1]).

• Complexity:

  • Runtime Complexity: O(C(n, k)), where C(n, k) represents the binomial coefficient "n choose k". This is because, in the worst case, we will explore all possible combinations.
  • Storage Complexity: O(k), primarily due to the depth of the recursion stack and the space used to store the current combination being built.

Code

    def combine(n: int, k: int) -> list[list[int]]:
    """
    Generates all possible combinations of k numbers chosen from the range [1, n].

    Args:
        n: The upper bound of the range (inclusive).
        k: The number of elements to choose in each combination.

    Returns:
        A list of lists, where each inner list represents a combination.
    """

    result = []

    def backtrack(start, combination):
        if len(combination) == k:
            result.append(combination.copy())  # Append a copy to avoid modification issues
            return

        for i in range(start, n + 1):
            combination.append(i)
            backtrack(i + 1, combination)  # Explore combinations starting from the next number
            combination.pop()  # Backtrack: remove the last added number to explore other paths

    backtrack(1, [])
    return result