<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 define subproblem DP(i, j) to be the maximum size of the square with bottom right corner to be at i, j
        # Consider following situation:
        # ? ? ?
        # ? ? ?
        # ? ? 1
        # DP(i, j) is dependent on DP(i, j-1), DP(i-1, j), DP(i-1, j-1)
        # if all three is 1, then DP(i, j) is 2. If all three is 2, then DP(i, j) is 3.
        # Basically DP(i, j) is the minimum of these three values, + 1.

        memory = {}
        
        def topdown(i, j):
            if (i, j) in memory:
                return memory[(i, j)]

            if matrix[i][j] == '0':
                memory[(i, j)] = 0
                return 0

            else:
                if i == 0 or j == 0:
                    memory[(i, j)] = 1
                    return 1
                dependence = min(topdown(i-1, j), topdown(i, j-1), topdown(i-1, j-1))
                memory[(i, j)] = dependence + 1
                return memory[(i, j)]
        
        ans = 0
        for i in range(len(matrix)):
            for j in range(len(matrix[0])):
                ans = max(ans, topdown(i, j) ** 2)
        return ans
58625