Largest Magic Square

Largest Magic Square - Problem

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

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

Input & Output

Example 1 — Classic Magic Square

$ Input: grid = [[7,1,4],[2,5,8],[6,9,3]]

› Output: 1

💡 Note: The 3x3 grid is not a magic square because the row sums are different: 7+1+4=12, 2+5+8=15, 6+9+3=18. Since the rows don't have equal sums, this cannot be a magic square. The largest magic square is any 1x1 cell, so the answer is 1.

Example 2 — Contains 2x2 Magic Square

$ Input: grid = [[1,1,1],[1,1,1],[1,1,1]]

› Output: 3

💡 Note: All elements are 1, so any square subgrid will have equal row, column, and diagonal sums. The entire 3x3 grid is a magic square with all sums equal to 3.

Example 3 — Only 1x1 Magic Squares

$ Input: grid = [[1,2],[3,4]]

› Output: 1

💡 Note: The 2x2 grid has row sums [3,7], column sums [4,6], and diagonal sums [5,5]. Since not all sums are equal, the largest magic square is 1x1 with size 1.

Constraints

m == grid.length
n == grid[i].length
1 ≤ m, n ≤ 50
1 ≤ grid[i][j] ≤ 10⁶

Visualization

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Asked in

G Google 25 M Microsoft 20 a Amazon 15

The optimal approach uses prefix sums to enable O(1) range sum queries when validating magic squares. For each position and size, we check if all row sums, column sums, and diagonal sums are equal. Time: O(m²n²), Space: O(mn).

Common Approaches

✓ Frequency Count

⏱️ Time: N/A Space: N/A

Brute Force

⏱️ Time: O(m²n²k²) Space: O(1)

For each possible top-left corner and size, calculate all row sums, column sums, and diagonal sums by iterating through elements. Check if all sums are equal.

Prefix Sum Optimization

⏱️ Time: O(m²n²) Space: O(mn)

Build prefix sum arrays to quickly calculate any rectangular region sum. This eliminates the need to repeatedly sum rows and columns when checking magic squares.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

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Smart Actions

💡 Explanation

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