Question

474. Ones and Zeroes

中文文档

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

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

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]

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};
  }
};

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];
}