# [LeetCode]#1688. Count of Matches in Tournament

**Environment: Python 3.7**

**Key technique: logic**

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.

# Analysis:

- n team will get (n-1) matches in examples in Ex 1 and Ex 2.
- Namely, we can output as answer.

**Solution:**

`class Solution:`

def numberOfMatches(self, n: int) -> int:

return n-1

**Submissions:**