<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])
        rows = []
        cols = [[0] * n for _ in range(m)]
        for r in range(m):
            temp = [0] * n
            for c in range(n):
                if matrix[r][c] != "0":
                    if c > 0:
                        temp[c] = temp[c-1]+1
                    else:
                        temp[c] = 1
            rows.append(temp)

        for c in range(n):
            curr = 0
            for r in range(m):
                if matrix[r][c] != "0":
                    if r > 0:
                        cols[r][c] = cols[r-1][c]+1
                    else:
                        cols[r][c] = 1
        visited = set()
        print(rows)
        print(cols)
        ans = 0
        for i in range(m):
            for j in range(n):
                if (i, j) not in visited and matrix[i][j] == "1":
                    length = 1
                    visited.add((i, j))
                    ti, tj = i+1, j+1
                    while ti < m and tj < n and min(rows[ti][tj], cols[ti][tj]) >= (length+1):
                        length += 1
                        ti += 1
                        tj += 1
                    ans = max(ans, length**2) 
        return ans



                    
                

        
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