Problem

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.

Examples

Example 1

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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

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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

Solution

Method 1 – Simulation or Mathematical Formula

Intuition

Each match eliminates one team. To determine the winner, all teams except one must be eliminated, so the total number of matches is always n - 1.

Approach

  1. Start with n teams.
  2. In each round, if n is even, play n/2 matches; if n is odd, play (n-1)/2 matches and one team advances.
  3. Continue until one team remains.
  4. The total number of matches is n - 1 (since each match eliminates one team).

Code

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class Solution {
public:
    int numberOfMatches(int n) {
        return n - 1;
    }
};
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func NumberOfMatches(n int) int {
    return n - 1
}
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class Solution {
    public int numberOfMatches(int n) {
        return n - 1;
    }
}
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class Solution {
    fun numberOfMatches(n: Int): Int {
        return n - 1
    }
}
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class Solution:
    def numberOfMatches(self, n: int) -> int:
        return n - 1
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impl Solution {
    pub fn number_of_matches(n: i32) -> i32 {
        n - 1
    }
}
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class Solution {
    numberOfMatches(n: number): number {
        return n - 1;
    }
}

Complexity

  • ⏰ Time complexity: O(1), since the answer is computed directly.
  • 🧺 Space complexity: O(1), since no extra space is used.