<p>Given a <code>m * n</code> matrix of ones and zeros, return how many <strong>square</strong> submatrices have all ones.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> matrix =
[
&nbsp; [0,1,1,1],
&nbsp; [1,1,1,1],
&nbsp; [0,1,1,1]
]
<strong>Output:</strong> 15
<strong>Explanation:</strong> 
There are <strong>10</strong> squares of side 1.
There are <strong>4</strong> squares of side 2.
There is  <strong>1</strong> square of side 3.
Total number of squares = 10 + 4 + 1 = <strong>15</strong>.
</pre>

<p><strong class="example">Example 2:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= arr.length&nbsp;&lt;= 300</code></li>
	<li><code>1 &lt;= arr[0].length&nbsp;&lt;= 300</code></li>
	<li><code>0 &lt;= arr[i][j] &lt;= 1</code></li>
</ul>

class Solution:
    def countSquares(self, matrix: List[List[int]]) -> int:
        for i in range(len(matrix)):
            for j in range(1, len(matrix[0])):
                matrix[i][j] += matrix[i][j-1]
        for i in range(1, len(matrix)):
            for j in range(len(matrix[0])):
                matrix[i][j] += matrix[i-1][j]
        m = len(matrix)
        n = len(matrix[0])
        tot = 0
        for sq in range(1, min(m,n) + 1):
            # ptot = tot
            for i in range(sq-1, m):
                for j in range(sq-1, n):
                    q1,q2,q3,q4 = matrix[i][j],0,0,0
                    if i-sq >= 0:
                        q2 = matrix[i-sq][j]
                    if j-sq >= 0:
                        q3 = matrix[i][j-sq]
                    if i-sq >= 0 and j-sq >= 0:
                        q4 = matrix[i-sq][j-sq]
                    if q1 - q2 - q3 + q4 == sq*sq:
                        tot += 1
            # print(f"Number of squares of size {sq}: {tot - ptot}")
        return tot
20900