<p>A <code>k x k</code> <strong>magic square</strong> is a <code>k x k</code> grid filled with integers such that every row sum, every column sum, and both diagonal sums are <strong>all equal</strong>. The integers in the magic square <strong>do not have to be distinct</strong>. Every <code>1 x 1</code> grid is trivially a <strong>magic square</strong>.</p>

<p>Given an <code>m x n</code> integer <code>grid</code>, return <em>the <strong>size</strong> (i.e., the side length </em><code>k</code><em>) of the <strong>largest magic square</strong> that can be found within this grid</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/29/magicsquare-grid.jpg" style="width: 413px; height: 335px;" />
<pre>
<strong>Input:</strong> grid = [[7,1,4,5,6],[2,5,1,6,4],[1,5,4,3,2],[1,2,7,3,4]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The largest magic square has a size of 3.
Every row sum, column sum, and diagonal sum of this magic square is equal to 12.
- Row sums: 5+1+6 = 5+4+3 = 2+7+3 = 12
- Column sums: 5+5+2 = 1+4+7 = 6+3+3 = 12
- Diagonal sums: 5+4+3 = 6+4+2 = 12
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/29/magicsquare2-grid.jpg" style="width: 333px; height: 255px;" />
<pre>
<strong>Input:</strong> grid = [[5,1,3,1],[9,3,3,1],[1,3,3,8]]
<strong>Output:</strong> 2
</pre>

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

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 50</code></li>
	<li><code>1 &lt;= grid[i][j] &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution:
    def largestMagicSquare(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0]) # dimensions 
        rows = [[0]*(n+1) for _ in range(m)] # prefix sum along row
        cols = [[0]*n for _ in range(m+1)] # prefix sum along column
        
        for i in range(m):
            for j in range(n): 
                rows[i][j+1] = grid[i][j] + rows[i][j]
                cols[i+1][j] = grid[i][j] + cols[i][j]
        
        ans = 1
        for i in range(m): 
            for j in range(n): 
                diag = grid[i][j]
                for k in range(min(i, j)): 
                    ii, jj = i-k-1, j-k-1
                    diag += grid[ii][jj]
                    ss = {diag}
                    for r in range(ii, i+1): ss.add(rows[r][j+1] - rows[r][jj])
                    for c in range(jj, j+1): ss.add(cols[i+1][c] - cols[ii][c])
                    ss.add(sum(grid[ii+kk][j-kk] for kk in range(k+2))) # anti-diagonal
                    if len(ss) == 1: ans = max(ans, k+2)
        return ans   
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