<p>Given an <code>m x n</code> binary <code>matrix</code> filled with <code>0</code>&#39;s and <code>1</code>&#39;s, <em>find the largest square containing only</em> <code>1</code>&#39;s <em>and return its area</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/26/max1grid.jpg" style="width: 400px; height: 319px;" />
<pre>
<strong>Input:</strong> matrix = [[&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;],[&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;],[&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;],[&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;]]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/26/max2grid.jpg" style="width: 165px; height: 165px;" />
<pre>
<strong>Input:</strong> matrix = [[&quot;0&quot;,&quot;1&quot;],[&quot;1&quot;,&quot;0&quot;]]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[&quot;0&quot;]]
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 300</code></li>
	<li><code>matrix[i][j]</code> is <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>

class Solution:
    def maximalSquare(self, matrix: List[List[str]]) -> int:
        n, m = len(matrix), len(matrix[0])
        # initiate a 2-D dp to store the maximal square area @ (i,j)
        dp = [[None] * m for _ in range(n)]
        
        # for first row and col, fill the dp with its corresponding matrix number
        for nCol in range(m):
            dp[0][nCol] = int(matrix[0][nCol])
        
        for nRow in range(n):
            dp[nRow][0] = int(matrix[nRow][0])
        
        if '1' in [max(row) for row in matrix]:
            maximalSquareArea = 1
        else:
            return 0 # matrix is all zeroes
            
        for i in range(1, n):
            for j in range(1, m):
        
                if int(matrix[i][j]) != 0:
                    # compute minimum of the immediate left, top, top-left cell value
                    adjCellsMin = min(dp[i-1][j], min(dp[i-1][j-1], dp[i][j-1]))
                    # update dp at current (i,j)
                    dp[i][j] = int((adjCellsMin ** 0.5 + int(matrix[i][j])) ** 2)
                    maximalSquareArea = max(maximalSquareArea, dp[i][j])
                else:
                    dp[i][j] = 0

        return maximalSquareArea
21775