<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])
        hor = [[0] * n for _ in range(m)]
        ver = [[0] * n for _ in range(m)]

        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    hor[i][j] = hor[i][j-1] + 1 if j > 0 else 1
                    ver[i][j] = ver[i-1][j] + 1 if i > 0 else 1

        max_size = 0
        for i in range(m-1, -1, -1):
            for j in range(n-1, -1, -1):
                small = min(hor[i][j], ver[i][j])
                while small > max_size:
                    if hor[i-small+1][j] >= small and ver[i][j-small+1] >= small:
                        max_size = small
                    small -= 1

        return max_size * max_size
16600