<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>

from typing import List

class Solution:
    def maximalSquare(self, m: List[List[str]]) -> int:
        if not m or not m[0]:
            return 0

        def solve(i, j, memo):
            if i >= len(m) or j >= len(m[0]):
                return 0
            if m[i][j] == '0':
                return 0
            if (i, j) in memo:
                return memo[(i, j)]

            right = solve(i, j + 1, memo)
            down = solve(i + 1, j, memo)
            diag = solve(i + 1, j + 1, memo)

            memo[(i, j)] = 1 + min(right, down, diag)
            return memo[(i, j)]

        memo = {}
        max_side = 0
        for i in range(len(m)):
            for j in range(len(m[0])):
                max_side = max(max_side, solve(i, j, memo))

        return max_side * max_side

36075