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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: Dynamic Programming

Thinking

Selecting the most strings under budgets of zeros and ones is \(0\)-\(1\) knapsack with two weights. Subset search is too large.

\(f[i][j][k]\) is the best count using the first \(i\) strings, at most \(j\) zeros and \(k\) ones. Skip copies the previous row; take adds \(1\) when the budgets allow.

Count \(0/1\) in the current string before filling the row so the transition uses that item's cost.

We define \(f[i][j][k]\) as the maximum number of strings that can be obtained from the first \(i\) strings using \(j\) zeros and \(k\) ones. Initially, \(f[i][j][k]=0\), and the answer is \(f[sz][m][n]\), where \(sz\) is the length of the array \(strs\).

For \(f[i][j][k]\), we have two choices:

  • Do not select the \(i\)-th string, in which case \(f[i][j][k]=f[i-1][j][k]\);
  • Select the \(i\)-th string, in which case \(f[i][j][k]=f[i-1][j-a][k-b]+1\), where \(a\) and \(b\) are the number of zeros and ones in the \(i\)-th string, respectively.

We take the maximum of these two choices to obtain the value of \(f[i][j][k]\).

The final answer is \(f[sz][m][n]\).

The time complexity is \(O(sz \times m \times n)\), and the space complexity is \(O(sz \times m \times n)\), where \(sz\) is the length of the array \(strs\), and \(m\) and \(n\) are the upper limits on the number of zeros and ones, respectively.

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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]

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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;
    }
}

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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};
    }
};

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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
}

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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];
}

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impl Solution {
    pub fn find_max_form(strs: Vec<String>, m: i32, n: i32) -> i32 {
        let sz = strs.len();
        let m = m as usize;
        let n = n as usize;
        let mut f = vec![vec![vec![0; n + 1]; m + 1]; sz + 1];
        for i in 1..=sz {
            let a = strs[i - 1].chars().filter(|&c| c == '0').count();
            let b = strs[i - 1].len() - a;
            for j in 0..=m {
                for k in 0..=n {
                    f[i][j][k] = f[i - 1][j][k];
                    if j >= a && k >= b {
                        f[i][j][k] = f[i][j][k].max(f[i - 1][j - a][k - b] + 1);
                    }
                }
            }
        }
        f[sz][m][n] as i32
    }
}

Solution 2: Dynamic Programming (Space Optimization)

Thinking

Row \(i\) depends only on row \(i-1\). Update \(j\) and \(k\) downward and drop the first dimension. Space becomes \(O(mn)\), and a string cannot be taken twice.

We notice that the calculation of \(f[i][j][k]\) only depends on \(f[i-1][j][k]\) and \(f[i-1][j-a][k-b]\). Therefore, we can eliminate the first dimension and optimize the space complexity to \(O(m \times n)\).

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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]

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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;
    }
}

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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};
    }
};

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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
}

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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];
}

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impl Solution {
    pub fn find_max_form(strs: Vec<String>, m: i32, n: i32) -> i32 {
        let m = m as usize;
        let n = n as usize;
        let mut f = vec![vec![0; n + 1]; m + 1];

        for s in strs {
            let a = s.chars().filter(|&c| c == '0').count();
            let b = s.len() - a;
            for i in (a..=m).rev() {
                for j in (b..=n).rev() {
                    f[i][j] = f[i][j].max(f[i - a][j - b] + 1);
                }
            }
        }

        f[m][n]
    }
}

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