<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>

import numpy as np
class Solution:
    def countSquares(self, mat: List[List[int]]) -> int:
        m, n = len(mat), len(mat[0])
        dp = np.zeros((m, n))
        for r in range(1,m): 
            dp[r][0] = mat[r][0]
        for c in range(n): 
            dp[0][c] = mat[0][c]
        for r in range(1, m): 
            for c in range(1, n): 
                if mat[r][c] == 1: 
                    sl = min(dp[r-1][c], dp[r][c-1])
                    if mat[int(r - sl)][int(c - sl)] == 1: dp[r][c] = int(1+sl)
                    else: dp[r][c] = int(sl)
                    if dp[r][c] == 0: dp[r][c] += 1
        #print(dp)
        return(int(np.sum(dp)))
        #for i in dp: tot += sum(i)
        #return(int(tot))
31000