LeetCode 1895: Largest Magic Square Solution

Master LeetCode problem 1895 (Largest Magic Square), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1895. Largest Magic Square

Medium

Array Matrix Prefix Sum

Problem Explanation

Explanation

To find the largest magic square within the given grid, we can iterate over each cell in the grid and check if it can be the top-left corner of a magic square of some size. We start with the largest possible size k and iteratively reduce k until we find a valid magic square.

For each cell, we check if it's possible to form a magic square of size k with that cell as the top-left corner. To do this, we need to check the row sums, column sums, and diagonal sums of the k x k square starting from that cell.

The time complexity of this approach is O(m * n * min(m, n)^2) where m and n are the dimensions of the grid. The space complexity is O(1).

Solution Code

class Solution {
    public int largestMagicSquare(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int maxSize = Math.min(m, n);

        for (int k = maxSize; k >= 1; k--) {
            for (int i = 0; i <= m - k; i++) {
                for (int j = 0; j <= n - k; j++) {
                    if (isMagicSquare(grid, i, j, k)) {
                        return k;
                    }
                }
            }
        }

        return 1;
    }

    private boolean isMagicSquare(int[][] grid, int row, int col, int k) {
        int sum = 0;
        int sumDiag1 = 0;
        int sumDiag2 = 0;

        for (int i = 0; i < k; i++) {
            int rowSum = 0;
            int colSum = 0;
            sumDiag1 += grid[row + i][col + i];
            sumDiag2 += grid[row + i][col + k - 1 - i];

            for (int j = 0; j < k; j++) {
                rowSum += grid[row + i][col + j];
                colSum += grid[row + j][col + i];
            }

            if (i == 0) {
                sum = rowSum;
            }

            if (rowSum != sum || colSum != sum) {
                return false;
            }
        }

        return sum == sumDiag1 && sum == sumDiag2;
    }
}

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