<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();
        int m=grid[0].size();
        vector<vector<int>>col=grid;
        vector<vector<int>>row=grid;

        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(grid[i][j]==1)
                {
                    row[i][j]=(j>0)?1+row[i][j-1]:1;
                    col[i][j]=(i>0)?1+col[i-1][j]:1;
                }
                else
                {
                    row[i][j]=0;
                    col[i][j]=0;
                }
            }
        }

        /*
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                cout<<row[i][j]<<" ";
            }
            cout<<endl;
        }

        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                cout<<col[i][j]<<" ";
            }
            cout<<endl;
        }
        */

        int ans=0;

        for(int i=n-1;i>=0;i--)
        {
            for(int j=m-1;j>=0;j--)
            {
                int limit=min(row[i][j],col[i][j]);

                for(int min1=1;min1<=limit;min1++)
                {
                    if(i-(min1-1)>=0 and j-(min1-1)>=0 and row[i-(min1-1)][j]>=min1 and 
                    col[i][j-(min1-1)]>=min1)

                    ans=max(ans,min1);
                }
            }
        }

        return ans*ans;

        return 0;
    }
};
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