<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 __init__(self):
        self.cache = {}
        self.res = 0

    def countSquares(self, matrix: List[List[int]]) -> int:
        m, n = len(matrix), len(matrix[0])

        for r in range(m):
            self.cache[(r , 0)] = matrix[r][0]
        for c in range(n):
            self.cache[(0 , c)] = matrix[0][c]
        
        for r in range(1 , m):
            for c in range(1 , n):
                if matrix[r][c] == 1:
                    left = self.cache[(r , c - 1)]
                    above = self.cache[(r - 1 , c)]
                    diag = self.cache[(r - 1 , c - 1)]
                    self.cache[(r , c)] = 1 + min(left , above , diag)
                else:
                    self.cache[(r , c)] = 0

        for value in self.cache.values():
            self.res += value

        return self.res


                
28000