LeetCode #840 — MEDIUM

Magic Squares In Grid

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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?

Note: while a magic square can only contain numbers from 1 to 9, grid may contain numbers up to 15.

Example 1:

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.

Example 2:

Input: grid = [[8]]
Output: 0

Constraints:

row == grid.length
col == grid[i].length
1 <= row, col <= 10
0 <= grid[i][j] <= 15

Step 01

Brute Force Baseline

Problem summary: 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? Note: while a magic square can only contain numbers from 1 to 9, grid may contain numbers up to 15.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Math

Example 1

[[4,3,8,4],[9,5,1,9],[2,7,6,2]]

Example 2

[[8]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #840: Magic Squares In Grid
class Solution {
    private int m;
    private int n;
    private int[][] grid;

    public int numMagicSquaresInside(int[][] grid) {
        m = grid.length;
        n = grid[0].length;
        this.grid = grid;
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                ans += check(i, j);
            }
        }
        return ans;
    }

    private int check(int i, int j) {
        if (i + 3 > m || j + 3 > n) {
            return 0;
        }
        int[] cnt = new int[16];
        int[] row = new int[3];
        int[] col = new int[3];
        int a = 0, b = 0;
        for (int x = i; x < i + 3; ++x) {
            for (int y = j; y < j + 3; ++y) {
                int v = grid[x][y];
                if (v < 1 || v > 9 || ++cnt[v] > 1) {
                    return 0;
                }
                row[x - i] += v;
                col[y - j] += v;
                if (x - i == y - j) {
                    a += v;
                }
                if (x - i + y - j == 2) {
                    b += v;
                }
            }
        }
        if (a != b) {
            return 0;
        }
        for (int k = 0; k < 3; ++k) {
            if (row[k] != a || col[k] != a) {
                return 0;
            }
        }
        return 1;
    }
}

// Accepted solution for LeetCode #840: Magic Squares In Grid
func numMagicSquaresInside(grid [][]int) (ans int) {
	m, n := len(grid), len(grid[0])
	check := func(i, j int) int {
		if i+3 > m || j+3 > n {
			return 0
		}
		cnt := [16]int{}
		row := [3]int{}
		col := [3]int{}
		a, b := 0, 0
		for x := i; x < i+3; x++ {
			for y := j; y < j+3; y++ {
				v := grid[x][y]
				if v < 1 || v > 9 || cnt[v] > 0 {
					return 0
				}
				cnt[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 a != b {
			return 0
		}
		for k := 0; k < 3; k++ {
			if row[k] != a || col[k] != a {
				return 0
			}
		}
		return 1
	}
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			ans += check(i, j)
		}
	}
	return
}

# Accepted solution for LeetCode #840: Magic Squares In Grid
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))

// Accepted solution for LeetCode #840: Magic Squares In Grid
impl Solution {
    pub fn num_magic_squares_inside(grid: Vec<Vec<i32>>) -> i32 {
        let m = grid.len();
        let n = grid[0].len();
        let mut ans: i32 = 0;

        let check = |i: usize, j: usize, grid: &Vec<Vec<i32>>| -> i32 {
            if i + 3 > m || j + 3 > n {
                return 0;
            }

            let mut cnt = vec![0; 16];
            let mut row = vec![0; 3];
            let mut col = vec![0; 3];
            let mut a = 0;
            let mut b = 0;

            for x in i..i + 3 {
                for y in j..j + 3 {
                    let v = grid[x][y] as usize;
                    if v < 1 || v > 9 {
                        return 0;
                    }
                    cnt[v] += 1;
                    if cnt[v] > 1 {
                        return 0;
                    }

                    let vv = grid[x][y];
                    row[x - i] += vv;
                    col[y - j] += vv;

                    if x - i == y - j {
                        a += vv;
                    }
                    if x - i + y - j == 2 {
                        b += vv;
                    }
                }
            }

            if a != b {
                return 0;
            }

            for k in 0..3 {
                if row[k] != a || col[k] != a {
                    return 0;
                }
            }

            1
        };

        for i in 0..m {
            for j in 0..n {
                ans += check(i, j, &grid);
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #840: Magic Squares In Grid
function numMagicSquaresInside(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    const check = (i: number, j: number): number => {
        if (i + 3 > m || j + 3 > n) {
            return 0;
        }
        const cnt: number[] = Array(16).fill(0);
        const row: number[] = Array(3).fill(0);
        const col: number[] = Array(3).fill(0);
        let [a, b] = [0, 0];
        for (let x = i; x < i + 3; ++x) {
            for (let y = j; y < j + 3; ++y) {
                const v = grid[x][y];
                if (v < 1 || v > 9 || ++cnt[v] > 1) {
                    return 0;
                }
                row[x - i] += v;
                col[y - j] += v;
                if (x - i === y - j) {
                    a += v;
                }
                if (x - i === 2 - (y - j)) {
                    b += v;
                }
            }
        }
        if (a !== b) {
            return 0;
        }
        for (let k = 0; k < 3; ++k) {
            if (row[k] !== a || col[k] !== a) {
                return 0;
            }
        }
        return 1;
    };
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            ans += check(i, j);
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.