<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:
        self.matrix = matrix
        self.square_cache = {}
        self.hor_cache = {}
        self.ver_cache = {}

        output = 0
        for i in range(len(matrix)):
            for j in range(len(matrix[0])):
                curr = self.maximalSquareEndingAt(i, j)
                if curr > output:
                    output = curr
        return output**2

    def maximalSquareEndingAt(self, i, j):
        cache = self.square_cache
        matrix = self.matrix
        # print(cache)
        if (i, j) in cache:
            return cache[(i, j)]

        if i < 0 or j < 0:
            return 0
        val = matrix[i][j]
        if val == "0":
            return 0
        
        maximalSquareNeighbor = self.maximalSquareEndingAt(i-1, j-1)
        longestHorizontalLeft = self.longestHorizontalEndingAt(i, j-1)
        longestVerticalAbove = self.longestVerticalEndingAt(i-1, j)
        output = 1 + min(
            maximalSquareNeighbor,
            longestHorizontalLeft,
            longestVerticalAbove,
        )
        cache[(i, j)] = output
        return output 

    def longestHorizontalEndingAt(self, i, j):
        cache = self.hor_cache
        matrix = self.matrix
        if (i, j) in cache:
            return cache[(i, j)]
        if j < 0:
            return 0
        val = matrix[i][j]
        if val == "0":
            return 0
        
        output = 1 + self.longestHorizontalEndingAt(i, j-1)
        cache[(i, j)] = output
        return output

    def longestVerticalEndingAt(self, i, j):
        cache = self.ver_cache
        matrix = self.matrix
        if (i, j) in cache:
            return cache[(i, j)]
        if i < 0:
            return 0
        val = matrix[i][j]
        if val == "0":
            return 0
        
        output = 1 + self.longestVerticalEndingAt(i-1, j)
        cache[(i, j)] = output
        return output

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