Introduction

In this blog post, we will explore how to solve the LeetCode problem "1900" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 (player 1 is the first player in the row, player 2 is the second player in the row, etc.). The tournament consists of multiple rounds (starting from round number 1). 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. When the number of players is odd for the current round, the player in the middle automatically advances to the next round. For example, if the row consists of players 1, 2, 4, 6, 7 Player 1 competes against player 7. Player 2 competes against player 6. Player 4 automatically advances to the next round. After each round is over, the winners are lined back up in the row based on the original ordering assigned to them initially (ascending order). The players numbered firstPlayer and secondPlayer are the best in the tournament. They can win against any other player before they compete against each other. If any two other players compete against each other, either of them might win, and thus you may choose the outcome of this round. Given the integers n, firstPlayer, and secondPlayer, return an integer array containing two values, the earliest possible round number and the latest possible round number in which these two players will compete against each other, respectively. Example 1: Input: n = 11, firstPlayer = 2, secondPlayer = 4 Output: [3,4] Explanation: One possible scenario which leads to the earliest round number: First round: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 Second round: 2, 3, 4, 5, 6, 11 Third round: 2, 3, 4 One possible scenario which leads to the latest round number: First round: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 Second round: 1, 2, 3, 4, 5, 6 Third round: 1, 2, 4 Fourth round: 2, 4 Example 2: Input: n = 5, firstPlayer = 1, secondPlayer = 5 Output: [1,1] Explanation: The players numbered 1 and 5 compete in the first round. There is no way to make them compete in any other round. Constraints: 2 <= n <= 28 1 <= firstPlayer < secondPlayer <= n

Explanation

Here's the breakdown of the problem and the solution:

• High-Level Approach:

  • Earliest Round: Binary search on the possible round numbers. In each iteration, check if it's possible for both players to reach the current round without meeting each other earlier. This involves calculating how many players each must defeat to reach that round and checking if their positions allow them to avoid each other.
  • Latest Round: Simulate the tournament rounds, prioritizing the players with lower initial indices to win if they are about to face one of the target players. Keep simulating rounds until the two target players face each other or one of them is eliminated.

• Complexity:

  • Runtime Complexity: O(log(n)^2) - due to the binary search for the earliest round and O(n) for the simulation in the latest round. More precisely, the binary search is over the possible number of rounds which is log(n). And the simulation runs at most log(n) rounds.
  • Storage Complexity: O(1) - constant extra space is used.

Code

    def earliestAndLatest(n: int, firstPlayer: int, secondPlayer: int) -> list[int]:
    """
    Calculates the earliest and latest possible rounds in which two specific players compete in a tournament.

    Args:
        n: The total number of players in the tournament.
        firstPlayer: The index of the first player (1-based).
        secondPlayer: The index of the second player (1-based).

    Returns:
        A list containing the earliest and latest possible round numbers.
    """

    def can_reach_round(round_num: int) -> bool:
        """
        Checks if both players can reach the given round without meeting each other.
        """
        k = 1 << (round_num - 1)  # Number of players needed to win to reach this round

        a = min(firstPlayer, n - firstPlayer + 1)
        b = min(secondPlayer, n - secondPlayer + 1)

        return a <= k and b <= k and (a + b > k + 1) # Players must reach the round, and not meet before it.

    # Earliest Round (Binary Search)
    low = 1
    high = n.bit_length()  # Maximum number of rounds possible
    earliest = high

    while low <= high:
        mid = (low + high) // 2
        if can_reach_round(mid):
            earliest = mid
            high = mid - 1
        else:
            low = mid + 1

    # Latest Round (Simulation)
    latest = 0
    cur_n = n
    p1 = firstPlayer
    p2 = secondPlayer
    round_num = 0

    while True:
        round_num += 1
        if cur_n % 2 == 1:
            mid = (cur_n + 1) // 2
            if p1 > mid:
                p1 = p1 - (cur_n - mid)
            if p2 > mid:
                p2 = p2 - (cur_n - mid)

        if min(p1, cur_n - p1 + 1) + min(p2, cur_n - p2 + 1) <= cur_n + 1:
            latest = round_num
            break

        next_p1 = (p1 + (cur_n + 1) // 2 - 1) // 2
        next_p2 = (p2 + (cur_n + 1) // 2 - 1) // 2

        p1 = next_p1
        p2 = next_p2
        cur_n = (cur_n + 1) // 2

    return [earliest, latest]