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This approach involves treating the tournament rounds as iterative events where in each round, the number of matches is calculated, and non-advanced teams are eliminated until one team remains. This approach iterates over the number of teams until a winner is decided by subtracting the number of matches played from the total count of teams.
Time Complexity: O(log n) since in each step the number of teams is halved.
Space Complexity: O(1) since we only use a constant amount of additional memory.
1def number_of_matches(n):
2 matches = 0
3 while n > 1:
4 if n % 2 == 0:
5 matches += n // 2
6 n //= 2
7 else:
8 matches += (n - 1) // 2
9 n = (n - 1) // 2 + 1
10 return matches
11
12n = 7
13print(number_of_matches(n)) # Output: 6The Python solution is similar and follows the iterative process to compute matches by changing the number of teams until one remains. This approach efficiently uses integer division and conditional logic to handle odd and even cases.
Instead of simulating each round, this approach uses the observation that to find a winner among n teams, exactly n - 1 matches must be played. This is because each match eliminates exactly one team and n - 1 eliminations leave one champion.
Time Complexity: O(1)
Space Complexity: O(1)
1
int numberOfMatches(int n) {
return n - 1;
}
int main() {
int n = 7;
std::cout << numberOfMatches(n) << std::endl; // Output: 6
return 0;
}This solution in C++, like its C counterpart, uses a mathematical reasoning to compute the number of matches, knowing that each team except one must be eliminated.