Formatted question description: https://leetcode.ca/all/1688.html

# 1688. Count of Matches in Tournament (Easy)

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 = 7Output:6Explanation: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 = 14Output:13Explanation: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`

**Related Topics**:

Backtracking

## Solution 1.

```
// OJ: https://leetcode.com/problems/count-of-matches-in-tournament/
// Time: O(logN)
// Space: O(1)
class Solution {
public:
int numberOfMatches(int n) {
int ans = 0;
for (; n > 1; n = n / 2 + n % 2) ans += n / 2;
return ans;
}
};
```

## Solution 2.

Each match removes a loser. To get one winner from the `n`

teams, we need `n - 1`

matches.

```
// OJ: https://leetcode.com/problems/count-of-matches-in-tournament/
// Time: O(1)
// Space: O(1)
class Solution {
public:
int numberOfMatches(int n) {
return n - 1;
}
};
```

Java

```
class Solution {
public int numberOfMatches(int n) {
int matches = 0;
while (n > 1) {
matches += n / 2;
n = (n + 1) / 2;
}
return matches;
}
}
```