<p>Given a 2D <code>grid</code> of <code>0</code>s and <code>1</code>s, return the number of elements in&nbsp;the largest <strong>square</strong>&nbsp;subgrid that has all <code>1</code>s on its <strong>border</strong>, or <code>0</code> if such a subgrid&nbsp;doesn&#39;t exist in the <code>grid</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,1,1],[1,0,1],[1,1,1]]
<strong>Output:</strong> 9
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,1,0,0]]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= grid.length &lt;= 100</code></li>
	<li><code>1 &lt;= grid[0].length &lt;= 100</code></li>
	<li><code>grid[i][j]</code> is <code>0</code> or <code>1</code></li>
</ul>
class Solution:
    def largest1BorderedSquare(self, grid: List[List[int]]) -> int:
        m=len(grid)
        n=len(grid[0])
        dp=[[[0,0,0,0] for _ in range(n) ]for _ in range(m)]
        for i in range(m):
            for j in range(n):
                if grid[i][j]==0:continue
                if i==0 or j==0:
                    dp[i][j][0]=1
                    dp[i][j][1]=1
                else:
                    dp[i][j][0]=1+dp[i-1][j][0]
                    dp[i][j][1]=1+dp[i][j-1][1]
        for i in range(m-1,-1,-1):
            for j in range(n-1,-1,-1):
                if grid[i][j]==0:continue
                if i==m-1 or j==n-1: 
                    dp[i][j][2]=1
                    dp[i][j][3]=1
                else:
                    dp[i][j][2]=1+dp[i+1][j][2]
                    dp[i][j][3]=1+dp[i][j+1][3]

        res=0
        for i in range(m):
            for j in range(n):
                cur=res
                l=min(dp[i][j][2],dp[i][j][3])
                for k in range(res,l):
                    h=min(dp[i+k][j+k][0],dp[i+k][j+k][1])
                    if h>=k:cur=max(cur,k+1)
                res=cur
        return res**2


        
17800