Introduction

In this blog post, we will explore how to solve the LeetCode problem "464" 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

In the "100 game" two players take turns adding, to a running total, any integer from 1 to 10. The player who first causes the running total to reach or exceed 100 wins. What if we change the game so that players cannot re-use integers? For example, two players might take turns drawing from a common pool of numbers from 1 to 15 without replacement until they reach a total >= 100. Given two integers maxChoosableInteger and desiredTotal, return true if the first player to move can force a win, otherwise, return false. Assume both players play optimally. Example 1: Input: maxChoosableInteger = 10, desiredTotal = 11 Output: false Explanation: No matter which integer the first player choose, the first player will lose. The first player can choose an integer from 1 up to 10. If the first player choose 1, the second player can only choose integers from 2 up to 10. The second player will win by choosing 10 and get a total = 11, which is >= desiredTotal. Same with other integers chosen by the first player, the second player will always win. Example 2: Input: maxChoosableInteger = 10, desiredTotal = 0 Output: true Example 3: Input: maxChoosableInteger = 10, desiredTotal = 1 Output: true Constraints: 1 <= maxChoosableInteger <= 20 0 <= desiredTotal <= 300

Explanation

Here's the breakdown of the solution:

• Minimax with Memoization: The problem can be modeled as a minimax game where each player aims to maximize their chance of winning. We use memoization (dynamic programming) to avoid redundant calculations, as the game state repeats often.
• Representing Game State: The state of the game is primarily defined by the numbers that have been used so far. A bitmask (integer) is used to represent the set of available numbers. The desiredTotal is also part of the game state.

• Base Cases: If the desiredTotal is already non-positive, the first player wins. If all numbers have been used and desiredTotal is still positive, the first player loses.

• Runtime Complexity: O(2^{maxChoosableInteger} maxChoosableInteger), where the 2^{maxChoosableInteger} factor comes from the number of possible states (represented by the bitmask), and the maxChoosableInteger factor comes from trying each available integer from 1 to maxChoosableInteger in the worst case.

• Storage Complexity: O(2^{maxChoosableInteger}) due to the memoization dictionary storing results for each possible game state.

Code

    def canIWin(maxChoosableInteger: int, desiredTotal: int) -> bool:
    """
    Determines if the first player can force a win in the described game.

    Args:
        maxChoosableInteger: The largest integer that can be chosen.
        desiredTotal: The target total that needs to be reached or exceeded.

    Returns:
        True if the first player can force a win, False otherwise.
    """

    if (maxChoosableInteger * (maxChoosableInteger + 1)) // 2 < desiredTotal:
        return False
    if desiredTotal <= 0:
        return True

    memo = {}

    def can_win(used_numbers, remaining_total):
        if (used_numbers, remaining_total) in memo:
            return memo[(used_numbers, remaining_total)]

        if remaining_total <= 0:
            return False

        for i in range(1, maxChoosableInteger + 1):
            if not (used_numbers & (1 << i)):
                if not can_win(used_numbers | (1 << i), remaining_total - i):
                    memo[(used_numbers, remaining_total)] = True
                    return True

        memo[(used_numbers, remaining_total)] = False
        return False

    return can_win(0, desiredTotal)