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The Earliest and Latest Rounds Where Players Compete

Number: 2028

Difficulty: Hard

Paid? No

Companies: Google

Problem Description

Given n players standing in a row and numbered from 1 to n, the tournament pairs the iᵗʰ player from the start with the iᵗʰ player from the end in each round. When there is an odd number of players, the middle player automatically advances. Two special players (firstPlayer and secondPlayer) always defeat any other opponent until they face each other. By choosing the outcomes for all non-special matches arbitrarily, determine the earliest and the latest rounds in which these two players can finally compete.

Key Insights

  • In every round, players are paired from the front and back of the current sequence.
  • Winners are reassembled in their original order to form the lineup for the next round.
  • The two special players can win against any other opponents until they meet.
  • Simulation of rounds requires tracking the positions of the special players across rounds.
  • A recursive (or dynamic programming) approach with memoization can explore all possible outcomes for matches that do not involve the two special players.
  • The goal is to combine these simulation results to extract the minimum (earliest) and maximum (latest) round in which the two players can face off.

Space and Time Complexity

Time Complexity: The recursion explores states defined by the positions of the two players among the remaining players. Given the small constraint (n ≤ 28), the state space is limited and the overall time complexity is manageable, often exponential in the worst case but practical with memoization. Space Complexity: O(n * n) due to the memoization storing states based on positions and remaining players.

Solution

We solve the problem using a recursive approach with memoization. At each round, determine the new positions of the two special players after pairing up the players. There are three possibilities:

  1. If the two players are paired in the current round, record the round number as the answer.
  2. Otherwise, simulate the round by determining the new positions of firstPlayer and secondPlayer among the winners after the round.
  3. When other players compete, choose the outcomes arbitrarily. Hence, for every valid configuration (keeping in mind that non-special matches can be directed as needed), explore both scenarios to compute the minimum round (earliest meeting) and maximum round (latest meeting) over all possibilities.

Key points include:

  • Reconstructing the order of players based on their original position after each round.
  • Recursively simulating the next round with updated positions for the two special players.
  • Using memoization to cache computed results and avoid recomputation.

Data structures:

  • A memoization map (dictionary) that stores a tuple key representing the current state (such as the positions of the two players and the remaining number of players) and maps to a pair of values: (earliest round, latest round).
  • Recursive functions to simulate rounds and update the positions of the special players.

Algorithm:

  • Define a recursive function that takes the current list (or positions) of players and the current round number.
  • If the players are matched in the current round, return the round number.
  • Otherwise, for every valid pairing outcome of non-special players, update the positions of firstPlayer and secondPlayer.
  • Compute the earliest and latest rounds over all simulation branches.
  • Use memoization to store these results to avoid recomputation.

The major trick is to correctly calculate the new positions after a round given that players are reassembled based on their original ordering. This challenge, combined with state branching due to arbitrary outcomes in matches not involving our special players, makes it suitable for a DP solution with recursion.

Code Solutions

# Python solution with line-by-line comments

def earliestAndLatest(n: int, firstPlayer: int, secondPlayer: int):
    from functools import lru_cache

    # Recursive function: players is a tuple of player numbers in the current round, round_num is current round number.
    @lru_cache(maxsize=None)
    def dfs(players, round_num):
        players = list(players)
        size = len(players)
        # Find indices (positions) of our special players in the current round.
        idx1 = players.index(firstPlayer)
        idx2 = players.index(secondPlayer)
        # If these two are paired, then their sum of indices equals size-1 and they are opposites.
        if idx1 + idx2 == size - 1:
            return (round_num, round_num)
        
        next_round = []
        # List to store all pairs of positions (earliest, latest) results from different branches.
        best_earliest = float('inf')
        best_latest = -float('inf')
        
        # Generate possible outcomes for matches.
        # We simulate the pairing by looping from both ends.
        left, right = 0, size - 1
        # List to hold all players that automatically advance (in case of odd number)
        winners_options = []
        def simulate(index, curr_winners, special_info):
            # index: current index in pairing simulation
            # curr_winners: current list of winners from earlier pairs
            # special_info: record if we've modified positions for special players if they appear
            if index >= size:
                # finished simulation of one branch, process next round
                new_players = sorted(curr_winners)  # maintain original order sorting (since players are labeled by original positions)
                res = dfs(tuple(new_players), round_num + 1)
                nonlocal best_earliest, best_latest
                best_earliest = min(best_earliest, res[0])
                best_latest = max(best_latest, res[1])
                return
            # If this is the middle element in odd number scenario, player automatically advances.
            if index == size - index - 1:
                curr_winners.append(players[index])
                simulate(index + 1, curr_winners, special_info)
                curr_winners.pop()
            else:
                # We are pairing players[index] with players[size - index - 1]
                a = players[index]
                b = players[size - index - 1]
                # If one of these players is special, they must win.
                # Outcome 1: a wins if special contender is a.
                if a == firstPlayer or a == secondPlayer:
                    curr_winners.append(a)
                    simulate(index + 1, curr_winners, special_info)
                    curr_winners.pop()
                # Outcome 2: b wins if b is special.
                elif b == firstPlayer or b == secondPlayer:
                    curr_winners.append(b)
                    simulate(index + 1, curr_winners, special_info)
                    curr_winners.pop()
                else:
                    # Neither player is special, we can choose the winner arbitrarily.
                    # Outcome: a wins.
                    curr_winners.append(a)
                    simulate(index + 1, curr_winners, special_info)
                    curr_winners.pop()
                    # Outcome: b wins.
                    curr_winners.append(b)
                    simulate(index + 1, curr_winners, special_info)
                    curr_winners.pop()
                    
        simulate(0, [], {})
        return (best_earliest, best_latest)
    
    # Initially players are in order 1,2,...,n.
    return dfs(tuple(range(1, n+1)), 1)

# Example usage:
print(earliestAndLatest(11, 2, 4))
print(earliestAndLatest(5, 1, 5))
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