Min Max Game - Leetcode Solution

The "Min Max Game" problem on Leetcode is a classic game theory problem that challenges you to use your strategic skills and intuition to come up with an optimal solution. The problem statement is as follows:

You are given an array of N integers, and two players take turns choosing one number each from the array until all the numbers have been chosen. Player 1 goes first, and the players alternate turns.

The game is played in such a way that player 1 always aims to maximize the sum of the numbers he has chosen, while player 2 always aims to minimize this sum.

Your task is to write a function that takes in the array and returns the maximum sum that player 1 can achieve, assuming that both players play optimally.

To solve this problem, you'll need to use a modified version of the minimax algorithm, which is a well-known algorithm in game theory. The idea is to analyze all possible moves and pick the move that maximizes the minimum gain of the opponent.

Here's how you can solve the Min Max Game problem on Leetcode:

1. Define a recursive function that takes in the array, the starting and ending indices, and a flag to indicate which player is currently playing. The base case is when the starting index is greater than the ending index, in which case you return 0.

2. If player 1 is currently playing, iterate through all possible moves by starting at the first position and choosing one number at a time. For each move, recursively call the function with the updated array and indices, and with player 2 as the current player. Take the maximum of all these values and return it as the final result.

3. If player 2 is currently playing, iterate through all possible moves by starting at the first position and choosing one number at a time. For each move, recursively call the function with the updated array and indices, and with player 1 as the current player. Take the minimum of all these values and return it as the final result.

4. Finally, call the recursive function with the given array, starting index 0, ending index N-1, and player 1 as the current player.

Here's the Python code that implements the above algorithm:

def minMaxGame(nums):
    def recursiveGame(nums, start, end, isMaxPlayer):
        if start > end:
            return 0

        if isMaxPlayer:
            best = float('-inf')
            for i in range(start, end + 1):
                best = max(best, recursiveGame(nums, start + 1, end, not isMaxPlayer) + nums[start])
            return best
        else:
            best = float('inf')
            for i in range(start, end + 1):
                best = min(best, recursiveGame(nums, start + 1, end, not isMaxPlayer) - nums[start])
            return best

    return recursiveGame(nums, 0, len(nums) - 1, True)

The time complexity of this algorithm is O(2^N), since in the worst case we have to make 2 binary choices N times. However, the use of memoization can significantly reduce this time complexity.

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