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 one 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

 

Example:

[0,1,1,1],
[1,1,1,1],
[0,1,1,1]

1. count single ones
2. count 2×2
3. count 3×3
4. ……

Goal: reduce n x n to (n-1)x(n-1) … to 2×2

Now scan 2×2 block, for each position(i,j) scan (i+1,j) (i,j+1) (i+1, j+1)
[0,1,1,1],
[1,1,1,1],
[0,1,1,1]

if the 2×2 scanning block contains 0, update (i,j) to 0
[0,1,1,1],
[0,1,1,1],
[0,1,1,1]

then we can reduce it by removing (ignoring) the last row and column.

[0,1,1],
[0,1,1]

Then walk through with the 2×2 scanning block again.

the counter could be updated when the whole 2×2 block is 1, OR count number of ones in the next round.

Version A, update counter during 2×2 scan

class Solution {
    public int countSquares(int[][] matrix) {
        int c = 0;
        
        int m  = matrix.length;
        int n  = matrix[0].length;
        
        for (int i = 0; i < m; i ++){
            for (int j = 0; j < n; j ++){
                if (matrix[i][j] == 1){
                    c++;
                } 
            }
        }
        
        
        while (m > 1 && n > 1){
            for (int i = 0; i < m-1; i ++){
                for (int j = 0; j < n-1; j ++){
                    if (  matrix[i][j]     == 0 
                       || matrix[i+1][j]   == 0
                       || matrix[i][j+1]   == 0
                       || matrix[i+1][j+1] == 0)
                    {
                        matrix[i][j] = 0;
                    }  else {
                        c++;
                    }
                }
            }
            // lower m and n (ignoring last row and column)
            m --;
            n --;
        }
        return c;
        
    }
}

 

B, a more compact version, update counter next round, count number of ones.
slightly slower for more loops

 

class Solution {
    public int countSquares(int[][] matrix) {
        int c = 0;
        
        int m  = matrix.length;
        int n  = matrix[0].length;
        
        
        
        while (m > 0 && n > 0){
            
            for (int i = 0; i < m; i ++){
                for (int j = 0; j < n; j ++){
                    if (matrix[i][j] == 1){
                        c++;
                    } 
                }
            }
            
            
            for (int i = 0; i < m-1; i ++){
                for (int j = 0; j < n-1; j ++){
                    if (matrix[i][j] == 0 
                       || matrix[i+1][j] == 0
                       || matrix[i][j+1] == 0
                       || matrix[i+1][j+1] == 0){
                        matrix[i][j] = 0;
                    }
                }
            }
            // lower m and n (ignoring last row and column)
            m --;
            n --;
        }
        return c;
        
    }
}

A: 86-110 ms
B: 160-170 ms