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
class Solution:
    def countSquares(self, matrix: List[List[int]]) -> int:
        m=len(matrix)
        n=len(matrix[0])
        ans=0
        for i in range(1,m):
            for j in range(1,n):
                if matrix[i][j]!=0 and matrix[i-1][j]!=0 and matrix[i][j-1]!=0 and matrix[i-1][j-1]!=0:
                    matrix[i][j]=min(matrix[i-1][j],matrix[i][j-1],matrix[i-1][j-1])+1
        for i in matrix:
            ans+=sum(i)
        return ans

class Solution:
    def countSquares(self, matrix: List[List[int]]) -> int:
        m=len(matrix)
        n=len(matrix[0])
        ans=0
        for i in range(1,m):
            for j in range(1,n):
                if matrix[i][j]!=0 and matrix[i-1][j]!=0 and matrix[i][j-1]!=0 and matrix[i-1][j-1]!=0:
                    matrix[i][j]=min(matrix[i-1][j],matrix[i][j-1],matrix[i-1][j-1])+1
        for i in matrix:
            ans+=sum(i)
        return ans
