<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, A: List[List[int]]) -> int:
        n, m, ans = len(A), len(A[0]), 0
        H, V = [row[:] for row in A], [row[:] for row in A]
        for i in range(n):
            for j in range(m):
                if i and H[i][j]: H[i][j] += H[i-1][j]
                if j and V[i][j]: V[i][j] += V[i][j-1]
        for i in range(n):
            for j in range(m):
                for k in range(min(H[i][j], V[i][j]), ans, -1):
                    if V[i-k+1][j] >= k and H[i][j-k+1] >= k:
                        ans = k
                        break
        return ans * ans
16800