<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:
        m = len(matrix)
        n = len(matrix[0])
        # maximal_len = m if m < n else n
        target_len = 1
        bad_starting_squares = set()
        def find_square(coord: tuple):
            searched_squares = set()
            if coord[0] + target_len > m:
                return None
            if coord[1] + target_len > n:
                return None
            for i in range(coord[0], coord[0] + target_len):
                for j in range(coord[1], coord[1] + target_len):
                    if matrix[i][j] == "0":
                        bad_starting_squares.add((i, j))
                        bad_starting_squares.add(coord)
                        return False
            return True
        def solution():
            nonlocal target_len
            found = False
            for i in range(m):
                if i + target_len > m:
                    break
                for j in range(n):
                    if j + target_len > n:
                        break
                    if (i, j) in bad_starting_squares:
                        continue
                    found = find_square((i, j))
                    if found:
                        target_len += 1
                        return solution()
                    if found is None:
                        bad_starting_squares.add((i, j))
                        break
            return (target_len - 1)**2
        
        return solution()

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