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1688. Count of Matches in Tournament

Description

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

Solutions

Solution 1: Quick Thinking

From the problem description, we know that there are $n$ teams in total. Each pairing will eliminate one team. Therefore, the number of pairings is equal to the number of teams eliminated, which is $n - 1$.

The time complexity is $O(1)$, and the space complexity is $O(1)$.

  • class Solution {
        public int numberOfMatches(int n) {
            return n - 1;
        }
    }
    
  • class Solution {
    public:
        int numberOfMatches(int n) {
            return n - 1;
        }
    };
    
  • class Solution:
        def numberOfMatches(self, n: int) -> int:
            return n - 1
    
    
  • func numberOfMatches(n int) int {
    	return n - 1
    }
    
  • function numberOfMatches(n: number): number {
        return n - 1;
    }
    
    
  • /**
     * @param {number} n
     * @return {number}
     */
    var numberOfMatches = function (n) {
        return n - 1;
    };
    
    

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