leetcode - 1895. Largest Magic Square

Description

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.

Example 1:

Input: grid = [[7,1,4,5,6],[2,5,1,6,4],[1,5,4,3,2],[1,2,7,3,4]]
Output: 3
Explanation: 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

Example 2:

Input: grid = [[5,1,3,1],[9,3,3,1],[1,3,3,8]]
Output: 2

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 50
1 <= grid[i][j] <= 10^6

Solution

Simulation, use a helper function to see if the row sums, column sums and diagonal sums are the same.

Time complexity: $o(m\ast n\ast n)$
Space complexity: $o(m\ast n)$

Code

class Solution:
    def largestMagicSquare(self, grid: List[List[int]]) -> int:
        def is_magic_square(x: int, y: int, size: int) -> bool:
            row_sums = [sum(grid[row][y: y + size]) for row in range(x, x + size)]
            col_sums = []
            for col in range(y, y + size):
                col_sums.append(0)
                for row in range(x, x + size):
                    col_sums[-1] += grid[row][col]
            diag_sums = [0, 0]
            for row in range(x, x + size):
                diag_sums[0] += grid[row][row - (x - y)]
                diag_sums[1] += grid[row][x + y + size - row - 1]
            return all(item == row_sums[0] for item in row_sums + col_sums + diag_sums)
        res = 1
        m, n = len(grid), len(grid[0])
        for i in range(m):
            for j in range(n):
                for size in range(2, 1 + min(m - i, n - j)):
                    if is_magic_square(i, j, size):
                        res = max(res, size)
        return res