<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, n = len(grid), len(grid[0])
        Table = [[[0,0] for _ in range(n)] for _ in range(m)]
        ans = 0
        for i in range(m-1, -1, -1):
            for j in range(n-1, -1, -1):
                if grid[i][j]:
                    Table[i][j][0] = 1 + (Table[i][j+1][0] if j+1<n and grid[i][j+1] else 0)
                    Table[i][j][1] = 1 + (Table[i+1][j][1] if i+1<m and grid[i+1][j] else 0)
                    for s in range(min(Table[i][j][0],Table[i][j][1]), -1 ,-1):
                        if Table[i][j+s-1][1] >= s and Table[i+s-1][j][0] >= s:
                            ans = max(ans, s)
                            break
        return ans**2
        
17700