algoadvance

You are given an integer maxChoosableInteger and a desired total. You and an opponent take turns picking numbers from 1 to maxChoosableInteger (inclusive) without replacement. The player who first reaches or exceeds the desired total wins. Assuming both players play optimally, return true if the first player can guarantee a win, otherwise return false.

Clarifying Questions

  1. Is input validation needed?
    • No, assume the inputs are valid integers within a reasonable range.
  2. Can maxChoosableInteger be larger than total?
    • Yes, maxChoosableInteger can be larger or smaller than total.
  3. Can numbers be picked more than once?
    • No, each number can only be picked once.
  4. What does it mean to play “optimally”?
    • Both players will make the move that maximizes their chances of winning.
  5. What are the constraints of the problem?
    • 1 <= maxChoosableInteger <= 20
    • 0 <= desiredTotal <= 300

Strategy

  1. Recursive Backtracking with Memorization:
    • We aim to simulate the game using recursion. A memoization table will be used to store already computed states to avoid re-computation.
    • Use a bitmask to keep track of the numbers that have been chosen so far.
  2. Base Cases:
    • If total <= 0, the player cannot win.
    • If maxChoosableInteger >= total, the first player can directly choose total and win.
  3. Recursive Function:
    • The function will simulate the game, deciding whether there’s a guaranteed win based on the current state.
    • Use memoization to store states (chosen numbers).
  4. Bitmasking:
    • To efficiently track chosen numbers, a bitmask will be used.
  5. Winning Strategy:
    • For each number not picked yet, recursively determine if choosing that number leads to a win.

Code

def canIWin(maxChoosableInteger, desiredTotal):
    if desiredTotal <= 0:
        return True
    if sum(range(1, maxChoosableInteger + 1)) < desiredTotal:
        return False

    memo = {}

    def can_win(remaining, chosen):
        if remaining <= 0:
            return False
        if chosen in memo:
            return memo[chosen]

        for i in range(1, maxChoosableInteger + 1):
            current_mask = 1 << i
            if chosen & current_mask == 0:
                if not can_win(remaining - i, chosen | current_mask):
                    memo[chosen] = True
                    return True

        memo[chosen] = False
        return False

    return can_win(desiredTotal, 0)

# Example usage:
maxChoosableInteger = 10
desiredTotal = 11
print(canIWin(maxChoosableInteger, desiredTotal))  # Output: False

Time Complexity

The time complexity of the solution can be analyzed as follows:

  1. Bitmask States: The total potential states we can have is 2^maxChoosableInteger, which is 2^20 in the worst case.
  2. Recursive Calls: Each call iterates over maxChoosableInteger choices.

Therefore, the time complexity is O(2^n * n), where n is maxChoosableInteger.

The space complexity is primarily dictated by the memoization table holding up to 2^n entries, making it O(2^n).

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