1688

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 = 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

Companies: Adobe, Yahoo

Related Topics:
Math, Simulation

Similar Questions:

• Count Distinct Numbers on Board (Easy)

Hints:

• Simulate the tournament as given in the statement.
• Be careful when handling odd integers.

Solution 1.

// OJ: https://leetcode.com/problems/count-of-matches-in-tournament/
// Author: github.com/lzl124631x
// 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/
// Author: github.com/lzl124631x
// Time: O(1)
// Space: O(1)
class Solution {
public:
    int numberOfMatches(int n) {
        return n - 1;
    }
};