Leetcode #840: Magic Squares In Grid
In this guide, we solve Leetcode #840 Magic Squares In Grid 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 3 x 3 magic square is a 3 x 3 grid filled with distinct numbers from 1 to 9 such that each row, column, and both diagonals all have the same sum. Given a row x col grid of integers, how many 3 x 3 magic square subgrids are there?
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Hash Table, Math, Matrix
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: grid = [[4,3,8,4],[9,5,1,9],[2,7,6,2]]
Output: 1
Explanation:
The following subgrid is a 3 x 3 magic square:
while this one is not:
In total, there is only one magic square inside the given grid.
Python Solution
class Solution:
def numMagicSquaresInside(self, grid: List[List[int]]) -> int:
def check(i: int, j: int) -> int:
if i + 3 > m or j + 3 > n:
return 0
s = set()
row = [0] * 3
col = [0] * 3
a = b = 0
for x in range(i, i + 3):
for y in range(j, j + 3):
v = grid[x][y]
if v < 1 or v > 9:
return 0
s.add(v)
row[x - i] += v
col[y - j] += v
if x - i == y - j:
a += v
if x - i == 2 - (y - j):
b += v
if len(s) != 9 or a != b:
return 0
if any(x != a for x in row) or any(x != a for x in col):
return 0
return 1
m, n = len(grid), len(grid[0])
return sum(check(i, j) for i in range(m) for j in range(n))
Complexity
The time complexity is , where and are the number of rows and columns of the matrix, respectively. 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.