<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:
        M = len(matrix)
        N = len(matrix[0])

        dp = [[0] * N for i in range(M)]

        dp[0][0] = 1 if matrix[0][0] == '1' else 0

        for m in range(1, M):
            dp[m][0] = 1 if matrix[m][0] == '1' else 0
        
        for n in range(1, N):
            dp[0][n] = 1 if matrix[0][n] == '1' else 0

        print(dp)

        for m in range(1, M):
            for n in range(1, N):
                if matrix[m][n] == '0':
                    continue

                dp[m][n] = 1            
                x = dp[m-1][n-1]
                if dp[m-1][n-1] > 0 and dp[m][n-1] > 0 and dp[m-1][n] > 0:
                    x = min(dp[m-1][n-1], dp[m][n-1], dp[m-1][n])
                    dp[m][n] = (x ** 0.5 + 1) ** 2
                    print(m, n, x, dp[m][n])

        print(dp)

        ret = -1
        for m in range(M):
            for n in range(N):
                ret = max(ret, dp[m][n])

        return int(ret)



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