class Solution:
    def lastStoneWeightII(self, stones: List[int]) -> int:
        
        @functools.cache  # Python functools自带记忆化单元
        def dfs(i, remain):
            '''remain: 前i个石头的表达式之和【可为负】'''
            if i == len(stones):            # 遍历完了全部的元素，递归中止
                return abs(remain)          # 返回绝对值【满足非负】
            
            pos = dfs(i+1, remain + stones[i])      # stones[i]前面添加'+'号
            neg = dfs(i+1, remain - stones[i])      # stones[i]前面添加'-'号
            return min(pos, neg)            # 返回两种情况中的最小值
        
        return dfs(0, 0)

class Solution:
    def findTargetSumWays(self, nums: List[int], target: int) -> int:
        s = sum(nums) - abs(target)
        if s < 0 or s % 2:
            return 0
        m = s // 2  # 背包容量

        @cache  # 缓存装饰器，避免重复计算 dfs 的结果（记忆化）
        def dfs(i: int, c: int) -> int:
            if i < 0:
                return 1 if c == 0 else 0
            if c < nums[i]:
                return dfs(i - 1, c)  # 只能不选
            return dfs(i - 1, c) + dfs(i - 1, c - nums[i])  # 不选 + 选
        return dfs(len(nums) - 1, m)

class Solution:
    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
        dp = [[0] * (n + 1) for _ in range(m + 1)]	# 默认初始化0
        # 遍历物品
        for str in strs:
            ones = str.count('1')
            zeros = str.count('0')
            # 遍历背包容量且从后向前遍历！
            for i in range(m, zeros - 1, -1):
                for j in range(n, ones - 1, -1):
                    dp[i][j] = max(dp[i][j], dp[i - zeros][j - ones] + 1)
        return dp[m][n]