Leetcode #1895: Largest Magic Square
In this guide, we solve Leetcode #1895 Largest Magic Square in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Matrix, Prefix Sum
Intuition
Range queries become simple once we precompute cumulative sums.
We can transform subarray conditions into prefix comparisons.
Approach
Compute prefix sums and use a map to find matching prefixes.
This avoids nested loops while keeping the logic clear.
Steps:
- Compute prefix sums.
- Use a map to find valid ranges.
- Update the answer.
Example
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
Python Solution
class Solution:
def largestMagicSquare(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
rowsum = [[0] * (n + 1) for _ in range(m + 1)]
colsum = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(1, m + 1):
for j in range(1, n + 1):
rowsum[i][j] = rowsum[i][j - 1] + grid[i - 1][j - 1]
colsum[i][j] = colsum[i - 1][j] + grid[i - 1][j - 1]
def check(x1, y1, x2, y2):
val = rowsum[x1 + 1][y2 + 1] - rowsum[x1 + 1][y1]
for i in range(x1 + 1, x2 + 1):
if rowsum[i + 1][y2 + 1] - rowsum[i + 1][y1] != val:
return False
for j in range(y1, y2 + 1):
if colsum[x2 + 1][j + 1] - colsum[x1][j + 1] != val:
return False
s, i, j = 0, x1, y1
while i <= x2:
s += grid[i][j]
i += 1
j += 1
if s != val:
return False
s, i, j = 0, x1, y2
while i <= x2:
s += grid[i][j]
i += 1
j -= 1
if s != val:
return False
return True
for k in range(min(m, n), 1, -1):
i = 0
while i + k - 1 < m:
j = 0
while j + k - 1 < n:
i2, j2 = i + k - 1, j + k - 1
if check(i, j, i2, j2):
return k
j += 1
i += 1
return 1
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.