<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:
        # keep track of length of chain of 1s below and right
        # get min of below chain, right chain, and below_right dp subresult
        # to make current dp result

        # dp_chain[i][j] = (below, right, square)
        # below and represents chain of 1s (i,j inclusive)
        # square represents dimensions of largest square
        # that can be made with (i,j) as top left corner

        dp = [[(0,0,0)] * len(matrix[0]) for _ in range(len(matrix))]
        ans = 0
        # set up base cases, very bottom and right cells
        for i in range(len(matrix)):
            if matrix[i][-1] == '1':
                dp[i][-1] = (1, 1, 1) # square dimensions
                ans = 1
        for j in range(len(matrix[0])):
            if matrix[-1][j] == '1':
                dp[-1][j] = (1, 1, 1)
                ans = 1
        

        for i in range(len(matrix) - 2, -1, -1):
            for j in range(len(matrix[0]) - 2, -1, -1):
                if matrix[i][j] == '0':
                    continue
                below = 1 + dp[i + 1][j][0]
                right = 1 + dp[i][j + 1][1]
                
                inner_square_length = dp[i + 1][j + 1][2]
                square_length = min(below, right, inner_square_length + 1)
                dp[i][j] = (below, right, square_length)
                ans = max(ans, square_length ** 2)
                print(i, j, square_length)

        return ans
26175