<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:
        nrow, ncol = len(matrix), len(matrix[0])
        
        # dp[(i, j)]: largest square's side where the top left vertex is (i, j)
        dp = {}

        def dfs(row, col):
            # cannot form a square
            if row == nrow or col == ncol or matrix[row][col] == '0':
                return 0
            
            # if subproblem has been computed
            if (row, col) in dp:
                return dp[(row, col)]

            # side of largest square
            side_diag = dfs(row + 1, col + 1) # diag sq
            side_h = dfs(row, col + 1) # horizontal sq (right)
            side_v = dfs(row + 1, col) # vertical sq (down)

            # already 1 square (1) + min side of these square can form a square
            dp[(row, col)] = 1 + min(side_diag, side_h, side_v)

            return dp[(row, col)]
        
        max_side = 0
        for row in range(nrow):
            for col in range(ncol):
                # no need to filter "1" as it will be terminated by line 10
                max_side = max(max_side, dfs(row, col))

        return max_side ** 2
55875