<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>

import numpy as np
class Solution:
    def maximalSquare(self, matrix: List[List[str]]) -> int:
        m, n = len(matrix), len(matrix[0])
        dp = np.zeros((m,n))
        for i in range(m):
            dp[i][0] = int(matrix[i][0])
        for j in range(n):
            dp[0][j] = int(matrix[0][j])
        res1 = np.max(dp[0, :])
        res2 = np.max(dp[:,0])
        res = max([res1, res2])
        for i in range(1, m):
            for j in range(1, n):
                if matrix[i][j]=='1':
                    side2 = min([dp[i-1][j-1], dp[i][j-1], dp[i-1][j]])
                    dp[i][j] = (np.sqrt(side2)+1)**2
                    res = max(res, dp[i][j])
        
        return int(res)












        # m, n = len(matrix), len(matrix[0])
        # dp = [[0]*(n+1) for i in range(m+1)]
        # res = 0
        # for i in range(m):
        #     for j in range(n):
        #         if matrix[i][j]=="1":
        #             dp[i+1][j+1]==1
        #             dp[i+1][j+1] = (np.sqrt(min([dp[i][j],dp[i+1][j], dp[i][j+1] ]))+1)**2
        #             res = max(res, dp[i+1][j+1])
        # #print(dp)
        # return int(res)


























        # for i in range(m):
        #     for j in range(n):
        #         if matrix[i][j]=='1':
        #             dp[i+1][j+1] = (sqrt(min([dp[i][j], dp[i+1][j], dp[i][j+1]]))+1)**2
        #             if dp[i+1][j+1]>res:
        #                 res = dp[i+1][j+1]
        # return int(res)
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        # m1 = np.zeros((m, n), dtype=int)
        # res = 0
        # for i in range(m):
        #     m1[i][0] = int(matrix[i][0])
        #     res = max(res, m1[i][0])
        # for j in range(n):
        #     m1[0][j] = int(matrix[0][j])
        #     res = max(res, m1[0][j])
        # #print(res)
        # for i in range(1, m):
        #     for j in range(1, n):
        #         if int(matrix[i][j]) == 1:
        #             res1 = min([m1[i][j-1], m1[i-1][j-1], m1[i-1][j]]) + 1
        #             m1[i][j] = res1
        #             res = max(res, res1)
        # print(m1)
        # return res*res

39375