Question

1277. Count Square Submatrices with All Ones

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Description

Given a m * n matrix of ones and zeros, return how many square submatrices have all ones.

 

Example 1:

Input: matrix =
[
  [0,1,1,1],
  [1,1,1,1],
  [0,1,1,1]
]
Output: 15
Explanation: 
There are 10 squares of side 1.
There are 4 squares of side 2.
There is  1 square of side 3.
Total number of squares = 10 + 4 + 1 = 15.

Example 2:

Input: matrix = 
[
  [1,0,1],
  [1,1,0],
  [1,1,0]
]
Output: 7
Explanation: 
There are 6 squares of side 1.  
There is 1 square of side 2. 
Total number of squares = 6 + 1 = 7.

 

Constraints:

• 1 <= arr.length <= 300
• 1 <= arr[0].length <= 300
• 0 <= arr[i][j] <= 1

Solutions

Solution 1

class Solution:
  def countSquares(self, matrix: List[List[int]]) -> int:
    m, n = len(matrix), len(matrix[0])
    f = [[0] * n for _ in range(m)]
    ans = 0
    for i, row in enumerate(matrix):
      for j, v in enumerate(row):
        if v == 0:
          continue
        if i == 0 or j == 0:
          f[i][j] = 1
        else:
          f[i][j] = min(f[i - 1][j - 1], f[i - 1][j], f[i][j - 1]) + 1
        ans += f[i][j]
    return ans

class Solution {
  public int countSquares(int[][] matrix) {
    int m = matrix.length;
    int n = matrix[0].length;
    int[][] f = new int[m][n];
    int ans = 0;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (matrix[i][j] == 0) {
          continue;
        }
        if (i == 0 || j == 0) {
          f[i][j] = 1;
        } else {
          f[i][j] = Math.min(f[i - 1][j - 1], Math.min(f[i - 1][j], f[i][j - 1])) + 1;
        }
        ans += f[i][j];
      }
    }
    return ans;
  }
}

class Solution {
public:
  int countSquares(vector<vector<int>>& matrix) {
    int m = matrix.size(), n = matrix[0].size();
    int ans = 0;
    vector<vector<int>> f(m, vector<int>(n));
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (matrix[i][j] == 0) continue;
        if (i == 0 || j == 0)
          f[i][j] = 1;
        else
          f[i][j] = min(f[i - 1][j - 1], min(f[i - 1][j], f[i][j - 1])) + 1;
        ans += f[i][j];
      }
    }
    return ans;
  }
};

func countSquares(matrix [][]int) int {
  m, n, ans := len(matrix), len(matrix[0]), 0
  f := make([][]int, m)
  for i := range f {
    f[i] = make([]int, n)
  }
  for i, row := range matrix {
    for j, v := range row {
      if v == 0 {
        continue
      }
      if i == 0 || j == 0 {
        f[i][j] = 1
      } else {
        f[i][j] = min(f[i-1][j-1], min(f[i-1][j], f[i][j-1])) + 1
      }
      ans += f[i][j]
    }
  }
  return ans
}