<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,n=len(matrix),len(matrix[0])
        matrix=[[int(matrix[j][i]) for i in range(n)] for j in range(m)]
        r1=[[0 for i in range(n)] for j in range(m)]
        for i in range(m):
            for j in range(n):
                if j==0:
                    r1[i][0]=matrix[i][0]
                else:
                    r1[i][j]=matrix[i][j]+r1[i][j-1]
        r2=[[0 for i in range(n)] for j in range(m)]
        for i in range(m):
            for j in range(n):
                if i==0:
                    r2[0][j]=r1[0][j]
                else:
                    r2[i][j]=r1[i][j]+r2[i-1][j]
        ans=0
        print(r2)
        for i in range(m):
            for j in range(n):
                for k in range(min(m-i,n-j)):
                    if i==0 and j==0:
                        if r2[k][k]==(k+1)*(k+1):
                            ans=max(ans,(k+1)*(k+1))
                    elif i==0:
                        if r2[i+k][j+k]-r2[i+k][j-1]==(k+1)*(k+1):
                            ans=max(ans,(k+1)*(k+1))
                    elif j==0:
                        if r2[i+k][j+k]-r2[i-1][j+k]==(k+1)*(k+1):
                            ans=max(ans,(k+1)*(k+1))
                    else:
                        if r2[i+k][j+k]+r2[i-1][j-1]-r2[i+k][j-1]-r2[i-1][j+k]==(k+1)*(k+1):
                            ans=max(ans,(k+1)*(k+1))
        return ans
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