Introduction

In this blog post, we will explore how to solve the LeetCode problem "1688" 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

You are given an integer n, the number of teams in a tournament that has strange rules: If the current number of teams is even, each team gets paired with another team. A total of n / 2 matches are played, and n / 2 teams advance to the next round. If the current number of teams is odd, one team randomly advances in the tournament, and the rest gets paired. A total of (n - 1) / 2 matches are played, and (n - 1) / 2 + 1 teams advance to the next round. Return the number of matches played in the tournament until a winner is decided. Example 1: Input: n = 7 Output: 6 Explanation: Details of the tournament: - 1st Round: Teams = 7, Matches = 3, and 4 teams advance. - 2nd Round: Teams = 4, Matches = 2, and 2 teams advance. - 3rd Round: Teams = 2, Matches = 1, and 1 team is declared the winner. Total number of matches = 3 + 2 + 1 = 6. Example 2: Input: n = 14 Output: 13 Explanation: Details of the tournament: - 1st Round: Teams = 14, Matches = 7, and 7 teams advance. - 2nd Round: Teams = 7, Matches = 3, and 4 teams advance. - 3rd Round: Teams = 4, Matches = 2, and 2 teams advance. - 4th Round: Teams = 2, Matches = 1, and 1 team is declared the winner. Total number of matches = 7 + 3 + 2 + 1 = 13. Constraints: 1 <= n <= 200

Explanation

Here's the solution:

• High-level approach: The core idea is to simulate the tournament round by round until a single winner is determined. In each round, calculate the number of matches and the number of advancing teams based on whether the current number of teams is even or odd, as described in the problem. Keep track of the total number of matches.

• The number of matches played is always n-1. This can be proved by induction. Base case n=1, the number of matches is 0, which is equal to 1-1. Now suppose for n=k, the number of matches played is k-1. Then for n=k+1, if k+1 is even, k+1 = 2m for some m. In this case, m matches are played in the first round, and m teams advance. The number of matches required for these m teams is m-1, by the inductive hypothesis. Therefore the total number of matches played is m + m -1 = 2m -1 = k+1 -1. Now, if k+1 is odd, k+1 = 2m+1 for some m. In this case, m matches are played in the first round, and m+1 teams advance. The number of matches required for these m+1 teams to determine the winner is m+1 -1 = m, by the inductive hypothesis. Therefore the total number of matches played is m + m = 2m = k+1-1. Therefore the induction holds.

• Complexity: O(1) time complexity, O(1) space complexity. The number of iterations of the while loop is at most log2(n) which, given the constraints 1 <= n <= 200, can be treated as O(1).

Code

    def numberOfMatches(n: int) -> int:
    """
    Calculates the number of matches played in the tournament until a winner is decided.

    Args:
        n: The number of teams in the tournament.

    Returns:
        The total number of matches played.
    """
    return n - 1