<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, mat: List[List[str]]) -> int:
        if not mat: return 0
        row = len(mat)
        col = len(mat[0])
      
        dp = [[0]*(col+1) for _ in range(row+1)]
        
        for i in range(1,row+1):
            for j in range(1,col+1):
                dp[i][j] = dp[i-1][j] + dp[i][j-1] - dp[i-1][j-1] + int(mat[i-1][j-1])
        print(dp)
        def possible(side):
            for i in range(side, row+1):
                for j in range(side, col+1):
                    cur = dp[i][j] - dp[i-side][j] - dp[i][j-side] + dp[i-side][j-side]
                    if cur == side*side:
                        return True
            return False
                 
                
        ans = 0
        l = 0
        r = min(row, col)
        
        while l <= r:
            mid = (l+r)//2
            if possible(mid):
                ans = max(mid, ans)
                l = mid + 1
            else:
                r = mid - 1
         
        return ans*ans
23425