Largest 1-Bordered Square
<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>
Array; Dynamic Programming; Matrix
class Solution:
    def largest1BorderedSquare(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        down = [[0] * n for _ in range(m)]
        right = [[0] * n for _ in range(m)]
        for i in range(m - 1, -1, -1):
            for j in range(n - 1, -1, -1):
                if grid[i][j]:
                    down[i][j] = down[i + 1][j] + 1 if i + 1 < m else 1
                    right[i][j] = right[i][j + 1] + 1 if j + 1 < n else 1
        for k in range(min(m, n), 0, -1):
            for i in range(m - k + 1):
                for j in range(n - k + 1):
                    if (
                        down[i][j] >= k
                        and right[i][j] >= k
                        and right[i + k - 1][j] >= k
                        and down[i][j + k - 1] >= k
                    ):
                        return k * k
        return 0
