474. Ones and Zeroes

Description

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

Solutions

Solution 1

Python3JavaC++GoTypeScript

class Solution:
    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
        sz = len(strs)
        f = [[[0] * (n + 1) for _ in range(m + 1)] for _ in range(sz + 1)]
        for i, s in enumerate(strs, 1):
            a, b = s.count("0"), s.count("1")
            for j in range(m + 1):
                for k in range(n + 1):
                    f[i][j][k] = f[i - 1][j][k]
                    if j >= a and k >= b:
                        f[i][j][k] = max(f[i][j][k], f[i - 1][j - a][k - b] + 1)
        return f[sz][m][n]

class Solution {
    public int findMaxForm(String[] strs, int m, int n) {
        int sz = strs.length;
        int[][][] f = new int[sz + 1][m + 1][n + 1];
        for (int i = 1; i <= sz; ++i) {
            int[] cnt = count(strs[i - 1]);
            for (int j = 0; j <= m; ++j) {
                for (int k = 0; k <= n; ++k) {
                    f[i][j][k] = f[i - 1][j][k];
                    if (j >= cnt[0] && k >= cnt[1]) {
                        f[i][j][k] = Math.max(f[i][j][k], f[i - 1][j - cnt[0]][k - cnt[1]] + 1);
                    }
                }
            }
        }
        return f[sz][m][n];
    }

    private int[] count(String s) {
        int[] cnt = new int[2];
        for (int i = 0; i < s.length(); ++i) {
            ++cnt[s.charAt(i) - '0'];
        }
        return cnt;
    }
}

class Solution {
public:
    int findMaxForm(vector<string>& strs, int m, int n) {
        int sz = strs.size();
        int f[sz + 1][m + 1][n + 1];
        memset(f, 0, sizeof(f));
        for (int i = 1; i <= sz; ++i) {
            auto [a, b] = count(strs[i - 1]);
            for (int j = 0; j <= m; ++j) {
                for (int k = 0; k <= n; ++k) {
                    f[i][j][k] = f[i - 1][j][k];
                    if (j >= a && k >= b) {
                        f[i][j][k] = max(f[i][j][k], f[i - 1][j - a][k - b] + 1);
                    }
                }
            }
        }
        return f[sz][m][n];
    }

    pair<int, int> count(string& s) {
        int a = count_if(s.begin(), s.end(), [](char c) { return c == '0'; });
        return {a, s.size() - a};
    }
};

func findMaxForm(strs []string, m int, n int) int {
    sz := len(strs)
    f := make([][][]int, sz+1)
    for i := range f {
        f[i] = make([][]int, m+1)
        for j := range f[i] {
            f[i][j] = make([]int, n+1)
        }
    }
    for i := 1; i <= sz; i++ {
        a, b := count(strs[i-1])
        for j := 0; j <= m; j++ {
            for k := 0; k <= n; k++ {
                f[i][j][k] = f[i-1][j][k]
                if j >= a && k >= b {
                    f[i][j][k] = max(f[i][j][k], f[i-1][j-a][k-b]+1)
                }
            }
        }
    }
    return f[sz][m][n]
}

func count(s string) (int, int) {
    a := strings.Count(s, "0")
    return a, len(s) - a
}

function findMaxForm(strs: string[], m: number, n: number): number {
    const sz = strs.length;
    const f = Array.from({ length: sz + 1 }, () =>
        Array.from({ length: m + 1 }, () => Array.from({ length: n + 1 }, () => 0)),
    );
    const count = (s: string): [number, number] => {
        let a = 0;
        for (const c of s) {
            a += c === '0' ? 1 : 0;
        }
        return [a, s.length - a];
    };
    for (let i = 1; i <= sz; ++i) {
        const [a, b] = count(strs[i - 1]);
        for (let j = 0; j <= m; ++j) {
            for (let k = 0; k <= n; ++k) {
                f[i][j][k] = f[i - 1][j][k];
                if (j >= a && k >= b) {
                    f[i][j][k] = Math.max(f[i][j][k], f[i - 1][j - a][k - b] + 1);
                }
            }
        }
    }
    return f[sz][m][n];
}

Solution 2