Desc

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

Code

class Solution:
    def numberOfMatches(self, n: int) -> int:
        ret = 0
        team = n
        while team > 1:
            if team % 2 == 0:
                ret += team // 2
                team //= 2
            else:
                ret += (team - 1) // 2
                team = (team - 1) // 2 + 1

        return ret

Code Desc

문제의 조건에서,

• n이 짝수 라면, n / 2 번의 매치가 발생하고, 남은 팀은 n / 2 개이다.
• n이 홀수 라면, (n - 1) / 2 번의 매치가 발생하고, 남은 팀은 (n - 1) / 2 + 1 개이다.

조건에 맞게 그대로 구현하면 된다.