# LeetCode 1688. Count of Matches in Tournament

## Description

https://leetcode.com/problems/count-of-matches-in-tournament/

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

Simply implement base on the rules.

## Python Solution

``````class Solution:
def numberOfMatches(self, n: int) -> int:
count = 0

while n > 1:
if n % 2 == 0:
count += n // 2
n = n // 2
else:
count += (n - 1) // 2
n = (n - 1) // 2 + 1

return count``````
• Time Complexity: O(N)
• Space Complexity: O(1)