<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:
        visited = {}
        max_square = 0
        def dfs(r, c):
            nonlocal max_square
            if r >= 0 and r < len(matrix) and c >= 0 and c < len(matrix[r]) and matrix[r][c] == "1":
                if (r, c) in visited:
                    return visited[(r, c)]

                min_size = math.inf
                for dr, dc in [[0, 1], [1, 0]]:
                    min_size = min(min_size, dfs(r + dr, c + dc))
                if matrix[r + min_size][c + min_size] == "1":
                    max_square = max(max_square, (min_size + 1) * (min_size + 1))
                    visited[(r, c)] = min_size + 1
                    return min_size + 1
                max_square = max(max_square, min_size if min_size else 1)
                visited[(r, c)] = min_size if min_size else 1
                return min_size if min_size else 1
            return 0

        max_square = 0
        for i in range(len(matrix)):
            for j in range(len(matrix[i])):
                if (i, j) not in visited and matrix[i][j] == "1":
                    dfs(i, j)

        return max_square
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