答案解析，可以转化为01背包问题，但因为有两个target，所以要用到三维dp数组。

python">from collections import Counter

class Solution:
    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
        nums0 = []
        nums1 = []
        length = len(strs)
        for s in strs:
            nums0.append(Counter(s)['0'])
            nums1.append(Counter(s)['1'])
        
        dp = [[[0] * (n+1) for _ in range(m+1)] for _ in range(length+1)]

        for i in range(1, length+1):
            for j in range(m+1):
                for k in range(n+1):
                    if j >= nums0[i-1] and k >= nums1[i-1]:
                        dp[i][j][k] = max(dp[i-1][j][k], dp[i-1][j-nums0[i-1]][k-nums1[i-1]]+1)
                    else:
                        dp[i][j][k] = dp[i-1][j][k]
        
        return dp[-1][-1][-1]

同其他01背包问题一样，可以将三维dp数组替换为二维滚动dp数组，同样要注意背包遍历要倒序以防止重复添加物品。

python">from collections import Counter

class Solution:
    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
        nums0 = []
        nums1 = []
        length = len(strs)
        for s in strs:
            nums0.append(Counter(s)['0'])
            nums1.append(Counter(s)['1'])
        
        dp = [[0] * (n+1) for _ in range(m+1)]

        for i in range(1, length+1):
            for j in range(m, -1, -1):
                for k in range(n, -1, -1):
                    if j >= nums0[i-1] and k >= nums1[i-1]:
                        dp[j][k] = max(dp[j][k], dp[j-nums0[i-1]][k-nums1[i-1]]+1)
                    else:
                        dp[j][k] = dp[j][k]
        
        return dp[-1][-1]