<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:
        memo = {} # (i, j):maxSquare
        def dfs(i, j):
            # out of bound or '0' in matrix, return 0
            # base case - last row, col
            # if in memo, return memo[pos]
            if i >= len(matrix) or j >= len(matrix[0]):
                return 0
            if i == len(matrix) - 1 or j == len(matrix[0]) - 1 or matrix[i][j] == '0':
                memo[(i, j)] = int(matrix[i][j])
                return memo[(i, j)]
            if (i, j) in memo:
                return memo[(i, j)]

            # i, j itself 0 -> 0
            # if one of the neighbor is '0' -> store and return the matrix[i][j] itself
            minSquare = float('inf')
            for dr, dc in [(1, 0), (0, 1), (1, 1)]:
                nei = dfs(i + dr, j + dc)
                if nei == 0:
                    memo[(i, j)] = int(matrix[i][j])
                    return memo[(i, j)]
                minSquare = min(minSquare, nei)
            # max square of (i, j) -> min(minSquare of neighbor) + 1
            memo[(i, j)] = minSquare + 1
            return memo[(i, j)]
        
        for r in range(len(matrix)):
            for c in range(len(matrix[0])):
                dfs(r, c) # being top left start point

        # iterate through memo to get max square
        maxSquare = 0
        for r, c in memo:
            maxSquare = max(maxSquare, memo[(r, c)])
        return maxSquare * maxSquare
43775