<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:
        if not matrix or not matrix[0]:
            return 0
        
        R, C = len(matrix), len(matrix[0])
        directions = [(1, 0), (0, 1), (1, 1)]  # Diagonal down and to the right
        max_length = 0
        memo = {}

        def dfs(r, c):
            if r >= R or r < 0 or c >= C or c < 0 or matrix[r][c] == "0":
                return 0
            if (r, c) in memo:
                return memo[(r, c)]

            # Calculate size of square
            down, right, diag = [dfs(r+dr, c+dc) for dr, dc in directions]
            length = min(down, right, diag) + 1
            memo[(r, c)] = length
            return memo[(r, c)]
        
        for r in range(R):
            for c in range(C):
                if matrix[r][c] == "1":
                    max_length = max(max_length, dfs(r, c))
        
        return max_length ** 2


# T: O(RC)
# S: O(RC)
        
37725