Solution Explanation

This problem asks to find the earliest and latest rounds in which two specific players (firstPlayer, secondPlayer) will compete against each other in a tournament. The tournament has n players, and in each round, players compete in pairs from the front and back of the line. The winners proceed to the next round.

The solution uses dynamic programming with memoization to efficiently explore the possible scenarios. The core idea is to recursively calculate the earliest and latest round numbers for any pair of players.

Approach

The solution utilizes a recursive function dp(l, r, k) with memoization:

• l: The position of firstPlayer from the front of the line in the current round.
• r: The position of secondPlayer from the end of the line in the current round.
• k: The total number of players in the current round.

The base cases are:

• If l == r, the players are in the same position, implying they compete in the current round (round 1).
• If l > r, we simply swap l and r since the order doesn't matter.

The recursive step considers all possible outcomes of the previous round. We iterate through possible positions (i, j) where i represents the position of firstPlayer and j represents the position of secondPlayer in the previous round. The constraints l + r - k // 2 <= i + j <= (k + 1) // 2 ensure that the positions i and j are valid given the number of players k in the previous round.

For each valid pair (i, j), we recursively call dp(i, j, (k + 1) // 2) to get the earliest and latest round numbers from the previous round. We then update the minimum earliest and maximum latest round numbers accordingly, adding 1 to account for the current round.

The initial call to dp is made with firstPlayer, n - secondPlayer + 1 (the position of secondPlayer from the end), and n (the total number of players).

Time and Space Complexity Analysis

• Time Complexity: The time complexity is difficult to express precisely because of the recursive nature and memoization. However, the memoization significantly reduces redundant calculations. In the worst case, without memoization, it would be exponential. With memoization, the time complexity is approximately O(n^3), where n is the number of players, due to the three nested loops in the recursive step (although the number of recursive calls is greatly reduced due to the memoization).

• Space Complexity: The space complexity is dominated by the memoization table. The size of the memoization table is proportional to the number of possible states, which is roughly O(n^3). Additionally, the recursion depth can be at most O(log n), leading to an additional space usage for the call stack. Therefore, the overall space complexity is O(n^3).

Python Code (with explanations)

import functools
import math
 
class Solution:
    def earliestAndLatest(self, n, firstPlayer, secondPlayer):
        @functools.lru_cache(None)  # Memoization decorator
        def dp(l, r, k):
            if l == r:
                return [1, 1]  # Base case: players compete in round 1
            if l > r:
                return dp(r, l, k)  # Swap l and r if necessary
 
            a = math.inf  # Initialize earliest round to infinity
            b = -math.inf  # Initialize latest round to negative infinity
 
            for i in range(1, l + 1):  # Iterate through possible previous positions for firstPlayer
                for j in range(l - i + 1, r - i + 1):  # Iterate through possible previous positions for secondPlayer
                    if not (l + r - k // 2 <= i + j <= (k + 1) // 2):
                        continue  #Skip invalid combinations
                    x, y = dp(i, j, (k + 1) // 2)  # Recursive call for previous round
                    a = min(a, x + 1)  # Update earliest round
                    b = max(b, y + 1)  # Update latest round
 
            return [a, b]
 
        return dp(firstPlayer, n - secondPlayer + 1, n)  # Initial call

The Python code directly implements the described dynamic programming approach with memoization using functools.lru_cache. The comments within the code further clarify each step. Other languages (like Java or C++) can implement this approach similarly, using their respective memoization techniques (e.g., HashMaps in Java).