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