<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])
        left = [[0] * (n + 1) for _ in range(m + 1)]
        up = [[0] * (n + 1) for _ in range(m + 1)]
        res = 0
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if grid[i - 1][j - 1]:
                    left[i][j] = left[i][j - 1] + 1
                    up[i][j] = up[i - 1][j] + 1
                    border = min(left[i][j], up[i][j])
                    while left[i - border + 1][j] < border or up[i][j - border + 1] < border:
                        border -= 1
                    res = max(res, border)
        return res * res
16700