Question

1504. Count Submatrices With All Ones

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Description

Given an m x n binary matrix mat, _return the number of submatrices that have all ones_.

 

Example 1:

Input: mat = [[1,0,1],[1,1,0],[1,1,0]]
Output: 13
Explanation: 
There are 6 rectangles of side 1x1.
There are 2 rectangles of side 1x2.
There are 3 rectangles of side 2x1.
There is 1 rectangle of side 2x2. 
There is 1 rectangle of side 3x1.
Total number of rectangles = 6 + 2 + 3 + 1 + 1 = 13.

Example 2:

Input: mat = [[0,1,1,0],[0,1,1,1],[1,1,1,0]]
Output: 24
Explanation: 
There are 8 rectangles of side 1x1.
There are 5 rectangles of side 1x2.
There are 2 rectangles of side 1x3. 
There are 4 rectangles of side 2x1.
There are 2 rectangles of side 2x2. 
There are 2 rectangles of side 3x1. 
There is 1 rectangle of side 3x2. 
Total number of rectangles = 8 + 5 + 2 + 4 + 2 + 2 + 1 = 24.

 

Constraints:

• 1 <= m, n <= 150
• mat[i][j] is either 0 or 1.

Solutions

Solution 1

class Solution:
  def numSubmat(self, mat: List[List[int]]) -> int:
    m, n = len(mat), len(mat[0])
    g = [[0] * n for _ in range(m)]
    for i in range(m):
      for j in range(n):
        if mat[i][j]:
          g[i][j] = 1 if j == 0 else 1 + g[i][j - 1]
    ans = 0
    for i in range(m):
      for j in range(n):
        col = inf
        for k in range(i, -1, -1):
          col = min(col, g[k][j])
          ans += col
    return ans

class Solution {
  public int numSubmat(int[][] mat) {
    int m = mat.length, n = mat[0].length;
    int[][] g = new int[m][n];
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (mat[i][j] == 1) {
          g[i][j] = j == 0 ? 1 : 1 + g[i][j - 1];
        }
      }
    }
    int ans = 0;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        int col = 1 << 30;
        for (int k = i; k >= 0 && col > 0; --k) {
          col = Math.min(col, g[k][j]);
          ans += col;
        }
      }
    }
    return ans;
  }
}

class Solution {
public:
  int numSubmat(vector<vector<int>>& mat) {
    int m = mat.size(), n = mat[0].size();
    vector<vector<int>> g(m, vector<int>(n));
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (mat[i][j] == 1) {
          g[i][j] = j == 0 ? 1 : 1 + g[i][j - 1];
        }
      }
    }
    int ans = 0;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        int col = 1 << 30;
        for (int k = i; k >= 0 && col > 0; --k) {
          col = min(col, g[k][j]);
          ans += col;
        }
      }
    }
    return ans;
  }
};

func numSubmat(mat [][]int) (ans int) {
  m, n := len(mat), len(mat[0])
  g := make([][]int, m)
  for i := range g {
    g[i] = make([]int, n)
    for j := range g[i] {
      if mat[i][j] == 1 {
        if j == 0 {
          g[i][j] = 1
        } else {
          g[i][j] = 1 + g[i][j-1]
        }
      }
    }
  }
  for i := range g {
    for j := range g[i] {
      col := 1 << 30
      for k := i; k >= 0 && col > 0; k-- {
        col = min(col, g[k][j])
        ans += col
      }
    }
  }
  return
}