'''

    In the "100 game," two players take turns adding, to a running total, any integer from 1..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 of 1..15 without replacement until
    they reach a total >= 100.

    Given an integer maxChoosableInteger and another integer desiredTotal, determine if the first player to move can
    force a win, assuming both players play optimally.

    You can always assume that maxChoosableInteger will not be larger than 20 and desiredTotal will not be larger than
    300.

    Link: https://leetcode.com/problems/can-i-win/#/description

    Example:

        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.

    Solution: None

    Source: None

'''


class Solution(object):
    def canIWin(self, maxChoosableInteger, desiredTotal):
        """
        :type maxChoosableInteger: int
        :type desiredTotal: int
        :rtype: bool
        """
        # generate a list to save all possible action/movement
        choosable = tuple(range(1, maxChoosableInteger + 1))
        if sum(choosable) < desiredTotal:
            return False
        self.cache = {}
        return self.dfs(choosable, desiredTotal)

    def dfs(self, choosable, total):
        # as explanation said, if the last value is large enough, just choose it, and  you will wine
        if choosable[-1] >= total:
            return True
        key = choosable
        if key in self.cache:
            return self.cache[key] # cache the previous solution to save time
        for i in range(len(choosable)):
            # ignore the i index. when i cannot be a solution
            if not self.dfs(choosable[:i] + choosable[i + 1:], total - choosable[i]):
                self.cache[key] = True
                return True
        self.cache[key] = False
        return False