You are given an array of binary strings strs and two integers m and n.

Return the size of the largest subset of strs such that there are at most m 0's and n 1's in the subset.

A set x is a subset of a set y if all elements of x are also elements of y.

Example 1:

Input: strs = ["10","0001","111001","1","0"], m = 5, n = 3
Output: 4
Explanation: The largest subset with at most 5 0's and 3 1's is {"10", "0001", "1", "0"}, so the answer is 4.
Other valid but smaller subsets include {"0001", "1"} and {"10", "1", "0"}.
{"111001"} is an invalid subset because it contains 4 1's, greater than the maximum of 3.

Example 2:

Input: strs = ["10","0","1"], m = 1, n = 1
Output: 2
Explanation: The largest subset is {"0", "1"}, so the answer is 2.

Constraints:

1 <= strs.length <= 600
1 <= strs[i].length <= 100
strs[i] consists only of digits '0' and '1'.
1 <= m, n <= 100

Solution

Top-Down DP

Time complexity : O(l*m*n) (l : size(strs))
Space complexity : O(l*m*n)

class Solution {
public:
    int findMaxForm(vector<string>& strs, int m, int n) {
        dp_.resize(size(strs), vector<vector<int>>(m+1, vector<int>(n+1)));
        return helper(strs, m, n, 0);
    }
private:
    vector<vector<vector<int>>> dp_;
    int helper(vector<string>& strs, int m, int n, int i) {
        if (i == size(strs) || m+n == 0) return 0;
        if (dp_[i][m][n]) return dp_[i][m][n];
        
        int zeros = count(begin(strs[i]), end(strs[i]), '0');
        int ones = size(strs[i]) - zeros;
        
        dp_[i][m][n] = helper(strs, m, n, i+1);
        if (zeros <= m && ones <= n)
            dp_[i][m][n] = max(dp_[i][m][n], 1+helper(strs, m-zeros, n-ones, i+1));
        
        return dp_[i][m][n];
    }
};

Bottom-Up DP

Time complexity : O(m*n)
Space complexity : O(m*n)

class Solution {
public:
    int findMaxForm(vector<string>& strs, int m, int n) {
        vector<vector<int>> dp(m+1, vector<int>(n+1));
        for (string s: strs) {
            int zeros = count(begin(s), end(s), '0');
            int ones = size(s) - zeros;
            for (int i=m; zeros<=i; --i) {
                for (int j=n; ones<=j; --j) {
                    dp[i][j] = max(dp[i][j],
                                1 + dp[i-zeros][j-ones]);
                }
            }
        }
        return dp[m][n];
    }
};