Welcome to Subscribe On Youtube

1277. Count Square Submatrices with All Ones

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

  • 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;
        }
    };
    
  • 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
    
    
  • 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
    }
    

All Problems

All Solutions