<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 {
public:
    int largest1BorderedSquare(vector<vector<int>>& grid) {
        int n = grid.size();
        if (n == 0) return 0;
        int m = grid[0].size();
        int maxSide = 0;

        // Auxiliary matrices to store the number of consecutive ones to the right and down
        vector<vector<int>> right(n, vector<int>(m, 0));
        vector<vector<int>> down(n, vector<int>(m, 0));

        // Preprocess the grid to fill right and down matrices
        for (int i = n - 1; i >= 0; --i) {
            for (int j = m - 1; j >= 0; --j) {
                if (grid[i][j] == 1) {
                    right[i][j] = (j + 1 < m) ? right[i][j + 1] + 1 : 1;
                    down[i][j] = (i + 1 < n) ? down[i + 1][j] + 1 : 1;
                } else {
                    right[i][j] = down[i][j] = 0;
                }
            }
        }

        // Find the largest square with ones on its borders
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                int small = min(right[i][j], down[i][j]);
                while (small > maxSide) {
                    if (right[i + small - 1][j] >= small && down[i][j + small - 1] >= small) {
                        maxSide = small;
                        break; // No need to check smaller sizes for this cell
                    }
                    --small;
                }
            }
        }

        return maxSide * maxSide; // Return the area of the largest square
    }
};

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