<p>A <code>3 x 3</code> <strong>magic square</strong> is a <code>3 x 3</code> grid filled with distinct numbers <strong>from </strong>1<strong> to </strong>9 such that each row, column, and both diagonals all have the same sum.</p>

<p>Given a <code>row x col</code> <code>grid</code> of integers, how many <code>3 x 3</code> magic square subgrids are there?</p>

<p>Note: while a magic square can only contain numbers from 1 to 9, <code>grid</code> may contain numbers up to 15.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/11/magic_main.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[4,3,8,4],[9,5,1,9],[2,7,6,2]]
<strong>Output:</strong> 1
<strong>Explanation: </strong>
The following subgrid is a 3 x 3 magic square:
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/11/magic_valid.jpg" style="width: 242px; height: 242px;" />
while this one is not:
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/11/magic_invalid.jpg" style="width: 242px; height: 242px;" />
In total, there is only one magic square inside the given grid.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[8]]
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>row == grid.length</code></li>
	<li><code>col == grid[i].length</code></li>
	<li><code>1 &lt;= row, col &lt;= 10</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;= 15</code></li>
</ul>

class Solution {
public:
    bool solve(vector<vector<int>>& grid, int startRow, int startCol) {
        vector<int> values;
        for (int i = startRow; i < startRow + 3; i++) {
            for (int j = startCol; j < startCol + 3; j++) {
                if (grid[i][j] < 1 || grid[i][j] > 9) {
                    return false;
                }
                values.push_back(grid[i][j]);
            }
        }
        sort(values.begin(), values.end());
        if (unique(values.begin(), values.end()) != values.end()) {
            return false; 
        }

        int sum1 = grid[startRow][startCol] + grid[startRow][startCol + 1] + grid[startRow][startCol + 2];
        int sum2 = grid[startRow + 1][startCol] + grid[startRow + 1][startCol + 1] + grid[startRow + 1][startCol + 2];
        int sum3 = grid[startRow + 2][startCol] + grid[startRow + 2][startCol + 1] + grid[startRow + 2][startCol + 2];


        int sum4 = grid[startRow][startCol] + grid[startRow + 1][startCol] + grid[startRow + 2][startCol];
        int sum5 = grid[startRow][startCol + 1] + grid[startRow + 1][startCol + 1] + grid[startRow + 2][startCol + 1];
        int sum6 = grid[startRow][startCol + 2] + grid[startRow + 1][startCol + 2] + grid[startRow + 2][startCol + 2];


        int sum7 = grid[startRow][startCol] + grid[startRow + 1][startCol + 1] + grid[startRow + 2][startCol + 2];
        int sum8 = grid[startRow][startCol + 2] + grid[startRow + 1][startCol + 1] + grid[startRow + 2][startCol];


        return (sum1 == 15 && sum2 == 15 && sum3 == 15 &&
                sum4 == 15 && sum5 == 15 && sum6 == 15 &&
                sum7 == 15 && sum8 == 15);
    }

    int numMagicSquaresInside(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int count = 0;

        // We only check 3x3 grids, so the loop boundaries must account for that
        for (int i = 0; i <= m - 3; i++) {
            for (int j = 0; j <= n - 3; j++) {
                if (solve(grid, i, j)) {
                    count++;
                }
            }
        }

        return count;
    }
};

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