Tournament Winner Rounds

Problem Description

We are given a tournament with n players arranged in a row, numbered from 1 to n. Each round involves players at the front and end of the row competing, with the winner advancing. The middle player advances automatically if the number of players is odd. After each round, winners are rearranged back in the row based on their original order. Given two best players, firstPlayer and secondPlayer, determine the earliest and latest rounds they can compete against each other.

Naive Approach

A brute-force approach involves simulating the tournament rounds and considering all possible outcomes of matches where firstPlayer and secondPlayer are not involved. This would involve recursively exploring different scenarios to find the earliest and latest rounds where the two players meet. However, this approach would be computationally expensive due to the exponential number of possible outcomes.

Optimal Approach

The problem can be solved using a binary search approach to determine the rounds each player can reach. We can calculate the maximum round a player can win by considering the number of players. After determining the maximum rounds each player can reach, we can compute the minimum and maximum rounds they can encounter each other.

1. Determine the Maximum Rounds: Calculate the maximum number of rounds a player can win to reach the final. This can be found by finding the largest k such that 2^k <= n. Alternatively, it can be calculated as floor(log2(n)). This determines the maximum number of rounds any player can participate in.
2. Calculate the Minimum Round: The earliest round they can meet is determined by how quickly they can be brought together. In the earliest possible scenario, all players besides the first and second advance as slowly as possible. Calculate how many players are removed from the beginning of the array for the first and second player. This difference dictates how many rounds each player needs to win before facing off.
3. Calculate the Maximum Round: The latest round they can meet is determined by delaying their meeting as long as possible. In this scenario, all players besides the first and second advance as quickly as possible. Then the meeting between first and second is delayed to the maximum extent.

Edge Cases

• If firstPlayer and secondPlayer are adjacent in the first round, they meet in the first round.
• If n is small (e.g., 2 or 3), the logic should still hold.
• Consider cases where firstPlayer is 1 and secondPlayer is n.

Code Example (Java)

class Solution {
    public int[] earliestAndLatest(int n, int firstPlayer, int secondPlayer) {
        return new int[]{minRound(n, firstPlayer, secondPlayer), maxRound(n, firstPlayer, secondPlayer)};
    }

    private int minRound(int n, int a, int b) {
        if (a > b) {
            int temp = a;
            a = b;
            b = temp;
        }
        if (a + b > n + 1) {
            b = n - a + 1;
        }
        if (a + b == n + 1) return 1;

        int round = 0;
        while (a != b) {
            a = (a + 1) / 2;
            b = (b + 1) / 2;
            round++;
        }
        return round + 1;
    }

    private int maxRound(int n, int a, int b) {
        if (a > b) {
            int temp = a;
            a = b;
            b = temp;
        }
        if (a + b > n + 1) {
            b = n - a + 1;
        }

        int round = 0;
        while (n > 1 && a != b) {
            a = (a + 1) / 2;
            b = (b + 1) / 2;
            n = (n + 1) / 2;
            round++;
        }
        return round + 1;
    }
}

Time and Space Complexity

• Time Complexity: O(log n) due to the iterative reduction of the problem size by half in each round.
• Space Complexity: O(1) as the solution uses a constant amount of extra space.