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

• class Solution {
public int findMaxForm(String[] strs, int m, int n) {
int[][] f = new int[m + 1][n + 1];
for (String s : strs) {
int[] cnt = count(s);
for (int i = m; i >= cnt[0]; --i) {
for (int j = n; j >= cnt[1]; --j) {
f[i][j] = Math.max(f[i][j], f[i - cnt[0]][j - cnt[1]] + 1);
}
}
}
return f[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 f[m + 1][n + 1];
memset(f, 0, sizeof(f));
for (auto& s : strs) {
auto [a, b] = count(s);
for (int i = m; i >= a; --i) {
for (int j = n; j >= b; --j) {
f[i][j] = max(f[i][j], f[i - a][j - b] + 1);
}
}
}
return f[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};
}
};

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


• func findMaxForm(strs []string, m int, n int) int {
f := make([][]int, m+1)
for i := range f {
f[i] = make([]int, n+1)
}
for _, s := range strs {
a, b := count(s)
for j := m; j >= a; j-- {
for k := n; k >= b; k-- {
f[j][k] = max(f[j][k], f[j-a][k-b]+1)
}
}
}
return f[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 f = 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 (const s of strs) {
const [a, b] = count(s);
for (let i = m; i >= a; --i) {
for (let j = n; j >= b; --j) {
f[i][j] = Math.max(f[i][j], f[i - a][j - b] + 1);
}
}
}
return f[m][n];
}