Count of Matches in Tournament

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Greedy Algorithms

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the input 'n' (number of teams) be zero or negative?
Is 'n' always an integer?
What is the maximum possible value for 'n'?
If 'n' is 1, should I return 0?
Are we only concerned with the total count of matches across all rounds, and not individual round match counts?

Brute Force Solution

Approach

In a tournament, teams are paired up to play matches. We want to figure out how many matches happen until only one team remains. The brute force strategy directly simulates each round of the tournament, counting the matches played in each round.

Here's how the algorithm would work step-by-step:

Start with the initial number of teams.
Figure out how many matches will be played in the current round by dividing the number of teams by two. If there's an odd number of teams, one team gets a bye (skips the round).
Count those matches towards the total matches played.
Calculate how many teams advance to the next round. This will be the number of matches played, plus one if there was a bye.
Update the number of teams to the number of teams advancing.
Repeat from step two until there is only one team left, meaning the tournament is over.
The total number of matches counted is the answer.

Code Implementation

def count_of_matches_in_tournament(number_of_teams):
    total_matches = 0
    teams_remaining = number_of_teams

    while teams_remaining > 1:
        # Calculate matches played in current round
        matches_in_round = teams_remaining // 2

        total_matches += matches_in_round

        # Calculate teams advancing to next round,
        # accounting for a possible bye.
        advancing_teams = matches_in_round
        if teams_remaining % 2 != 0:
            # One team gets a bye if odd number of teams
            advancing_teams += 1

        teams_remaining = advancing_teams

    return total_matches

Big(O) Analysis

Time Complexity

O(log n) – The number of teams is repeatedly halved in each round of the tournament. The while loop continues until the number of teams is one. The dominant operation is the division of the number of teams in each iteration. Since the input size n (number of teams) is halved in each step, the number of iterations required is logarithmic with respect to n. Therefore, the time complexity is O(log n).

Space Complexity

O(1) – The provided algorithm simulates a tournament by repeatedly calculating the number of matches and advancing teams. It uses a fixed number of variables to store the current number of teams and the total number of matches played. The space used does not depend on the initial number of teams (N) as it only involves storing a couple of integer values. Therefore, the auxiliary space complexity is constant.

Optimal Solution

Approach

The core idea is that in each match, one team loses, except for the final winner. So, to find the total matches, we need to figure out how many teams ultimately lose. This can be easily found by understanding the connection between the number of teams and the matches played.

Here's how the algorithm would work step-by-step:

Recognize that every team, except the winner, will lose exactly one match.
Therefore, the number of matches played is exactly equal to the number of teams that lost.
Since there is only one winner, the number of losing teams (and thus the number of matches) is just the total number of teams minus 1.

Code Implementation

def number_of_matches(number_of_teams):
    # Every team except the winner loses once.
    losing_teams = number_of_teams - 1

    # The number of matches is equal to losing teams.

    total_matches = losing_teams

    # Return the calculated number of matches.
    return total_matches

Big(O) Analysis

Time Complexity

O(1) – The solution directly calculates the number of matches by subtracting 1 from the number of teams (n). This involves only a single arithmetic operation. Therefore, the time complexity does not depend on the size of the input (n) and is constant, resulting in O(1) time complexity.

Space Complexity

O(1) – The provided explanation focuses on the relationship between the number of teams and the matches played. It doesn't describe any auxiliary data structures, temporary lists, hash maps, or recursion. The calculation involves only subtracting 1 from the input number of teams. Thus, no extra memory is used beyond the input itself, resulting in constant space complexity.

Edge Cases

Case	How to Handle
n = 0	Return 0, as there are no matches if no teams are present.
n = 1	Return 0, as a single team automatically wins without a match.
n = 2	Return 1, as there's one match between the two teams.
n is a large positive integer (e.g., 10^9)	The iterative or recursive solution should be efficient and avoid stack overflow; the formulaic solution (n-1) directly addresses the large n issue.
n is a negative number	Throw an exception or return an error code because the number of teams cannot be negative.
n is a non-integer number	Convert the number to an integer by truncating or rounding, documenting the approach taken.
Integer overflow during calculations when n is close to the maximum integer value.	Use a language that can handle large numbers, or if not, return a bigInteger type to avoid information loss.
n = 2^k for some integer k	The number of teams is a power of 2, so each round all teams can play, and this is a good test case for performance and correctness.

n = 0

How to Handle:

Return 0, as there are no matches if no teams are present.

n = 1

How to Handle:

Return 0, as a single team automatically wins without a match.

n = 2

How to Handle:

Return 1, as there's one match between the two teams.

n is a large positive integer (e.g., 10^9)

How to Handle:

The iterative or recursive solution should be efficient and avoid stack overflow; the formulaic solution (n-1) directly addresses the large n issue.

n is a negative number

How to Handle:

Throw an exception or return an error code because the number of teams cannot be negative.

n is a non-integer number

How to Handle:

Convert the number to an integer by truncating or rounding, documenting the approach taken.

Integer overflow during calculations when n is close to the maximum integer value.

How to Handle:

Use a language that can handle large numbers, or if not, return a bigInteger type to avoid information loss.

n = 2^k for some integer k

How to Handle:

The number of teams is a power of 2, so each round all teams can play, and this is a good test case for performance and correctness.