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The Earliest and Latest Rounds Where Players Compete

Difficulty: Hard

Problem Description

There is a tournament where n players are participating. The players are standing in a single row and are numbered from 1 to n based on their initial standing position. In each round, the ith player from the front of the row competes against the ith player from the end of the row, and the winner advances to the next round. If the number of players is odd, the player in the middle automatically advances. Given the integers n, firstPlayer, and secondPlayer, return an integer array containing the earliest and latest possible round number in which these two players will compete against each other.

Key Insights

  • Players compete in pairs from opposite ends of the row, narrowing down the number of players each round.
  • The order of players is restored after each round, making it essential to track their positions accurately.
  • The earliest round occurs when the two players face off as soon as possible, while the latest round occurs when they face off as late as possible.

Space and Time Complexity

Time Complexity: O(log n)
Space Complexity: O(1)

Solution

To determine when firstPlayer and secondPlayer will compete against each other, we can use a simulation approach. The primary strategy involves calculating their positions in each round until they compete.

  1. Simulate Rounds: In each round, calculate the new positions of players based on their current positions and the rules of the competition.
  2. Earliest Round Calculation: For the earliest round, we need to ensure that both players win their matches until they face each other.
  3. Latest Round Calculation: For the latest round, allow other players to win until the two players are the last remaining.

The players can be indexed using their initial positions, and we can continue halving the number of players until they are matched against each other.

Code Solutions

def earliest_and_latest_round(n: int, firstPlayer: int, secondPlayer: int) -> List[int]:
    def get_round(first, second):
        round_number = 0
        while first != second:
            round_number += 1
            first = (first + 1) // 2
            second = (n - second + 1) // 2
        return round_number

    earliest = get_round(firstPlayer, secondPlayer)
    
    # For the latest round
    latest = 0
    while firstPlayer // 2 != secondPlayer // 2:
        latest += 1
        firstPlayer = (firstPlayer + 1) // 2
        secondPlayer = (secondPlayer + 1) // 2

    return [earliest, latest + 1]  # +1 to include the final match
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