<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:
        def check_square(i, j, matrix, node_map):
            if i >= len(matrix) or j >= len(matrix[0]) or matrix[i][j] == "0":
                return 0
            if node_map.get((i,j)):
                return node_map.get((i,j))
            res = min(check_square(i, j+1, matrix, node_map), check_square(i+1, j, matrix, node_map), check_square(i+1, j+1, matrix, node_map))
            node_map[(i,j)] = res+1
            return node_map[(i,j)]



        def get_max_sqaure(i, j, matrix, node_map):
            max_length = 0
            for i in range(len(matrix)):
                for j in range(len(matrix[0])):
                    if matrix[i][j] == "1":
                        if node_map.get((i,j)):
                            length = node_map.get((i,j))
                        else:
                            length = check_square(i,j, matrix, node_map)
                        if length > max_length:
                            max_length = length
            return max_length**2

        node_map = {}
        return get_max_sqaure(0, 0, matrix, node_map)
        
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