<p>You are given an <code>n x n</code> integer matrix <code>grid</code>.</p>

<p>Generate an integer matrix <code>maxLocal</code> of size <code>(n - 2) x (n - 2)</code> such that:</p>

<ul>
	<li><code>maxLocal[i][j]</code> is equal to the <strong>largest</strong> value of the <code>3 x 3</code> matrix in <code>grid</code> centered around row <code>i + 1</code> and column <code>j + 1</code>.</li>
</ul>

<p>In other words, we want to find the largest value in every contiguous <code>3 x 3</code> matrix in <code>grid</code>.</p>

<p>Return <em>the generated matrix</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/06/21/ex1.png" style="width: 371px; height: 210px;" />
<pre>
<strong>Input:</strong> grid = [[9,9,8,1],[5,6,2,6],[8,2,6,4],[6,2,2,2]]
<strong>Output:</strong> [[9,9],[8,6]]
<strong>Explanation:</strong> The diagram above shows the original matrix and the generated matrix.
Notice that each value in the generated matrix corresponds to the largest value of a contiguous 3 x 3 matrix in grid.</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/07/02/ex2new2.png" style="width: 436px; height: 240px;" />
<pre>
<strong>Input:</strong> grid = [[1,1,1,1,1],[1,1,1,1,1],[1,1,2,1,1],[1,1,1,1,1],[1,1,1,1,1]]
<strong>Output:</strong> [[2,2,2],[2,2,2],[2,2,2]]
<strong>Explanation:</strong> Notice that the 2 is contained within every contiguous 3 x 3 matrix in grid.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length == grid[i].length</code></li>
	<li><code>3 &lt;= n &lt;= 100</code></li>
	<li><code>1 &lt;= grid[i][j] &lt;= 100</code></li>
</ul>

class Solution {
public:
    vector<vector<int>> largestLocal(vector<vector<int>>& grid) {

        //vector<vector<int>> answer(grid.size()-2, vector<int>(grid.size()-2));
        vector<vector<int>> answer;

        for(int i = 0; i < grid.size()-2; i++) {

            vector<int> arr;
         
            for(int j = 0; j < grid.size()-2; j++) {

                int greatest = 0;
                for(int k = i; k < i+3; k++) {
                    for(int l = j; l < j+3; l++) { 
                        if(grid[k][l] > greatest) {
                            greatest = grid[k][l];
                        }
                    }
                }
                arr.push_back(greatest);
            }
            answer.push_back(arr);
            
        }
        return answer;
    }
};
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