How do you solve Largest Magic Square on LeetCode?

Largest Magic Square (LeetCode #1895) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(m * n) space complexity. Key insight: Magic squares require equal sums across rows, columns, and diagonals.

What is the brute force solution for Largest Magic Square?

The brute force approach for Largest Magic Square is Brute Force, with O(m * n * min(m, n)) time complexity and O(1) space complexity. The brute force approach checks every possible square in the grid to see if it qualifies as a magic square. This is simple to understand but inefficient for larger grids. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Largest Magic Square?

Largest Magic Square uses the following concepts: Array, Matrix, Prefix Sum. The recommended patterns to study are: Dynamic Programming, Prefix Sums.

What are the interview tips for Largest Magic Square?

Clarify the definition of a magic square before starting. Discuss your thought process and approach before coding. Consider edge cases, such as very small grids.

What are common mistakes when solving Largest Magic Square?

Not checking all possible starting points for squares. Overlooking the need to check diagonal sums.

#1895

Largest Magic Square

Medium

ArrayMatrixPrefix SumDynamic ProgrammingPrefix Sums

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m * n * min(m, n))	O(m * n)
Space	O(1)	O(m * n)

💡

Intuition

Time O(m * n)Space O(m * n)

The optimal solution uses dynamic programming to build on previously computed results, allowing us to efficiently check for larger magic squares without recalculating sums from scratch.

⚙️

Algorithm

3 steps

1Step 1: Create a DP table to store the size of the largest magic square ending at each cell.
2Step 2: Iterate through the grid, and for each cell, check if it can be the bottom-right corner of a magic square.
3Step 3: Use previously computed values to determine the size of the magic square that can be formed.

solution.py13 lines

1def largestMagicSquare(grid):
2    m, n = len(grid), len(grid[0])
3    dp = [[0] * n for _ in range(m)]
4    max_size = 1
5    for i in range(m):
6        for j in range(n):
7            if i == 0 or j == 0:
8                dp[i][j] = 1
9            else:
10                if grid[i][j] == grid[i-1][j] == grid[i][j-1] == grid[i-1][j-1]:
11                    dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1
12            max_size = max(max_size, dp[i][j])
13    return max_size

ℹ

Complexity note: The time complexity is linear with respect to the number of cells in the grid, as we only iterate through the grid once and use previously computed results to determine the size of magic squares.

1Magic squares require equal sums across rows, columns, and diagonals.
2Dynamic programming can optimize checking for larger squares by leveraging smaller results.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.