Is Design Neighbor Sum Service easy or hard?

Design Neighbor Sum Service is classified as an Easy problem. The main challenge is recognizing that unique grid values allow constant-time coordinate lookup using a hash map, making neighbor calculations straightforward.

Design Neighbor Sum Service Python/Java solution

The Python and Java implementations both follow the same idea: build a HashMap or dictionary mapping values to their matrix coordinates, then compute neighbor sums using directional offsets. Each query runs in O(1) time after preprocessing.

How to solve Design Neighbor Sum Service in O(1)?

Store the coordinates of each grid value in a hash map during initialization. When a query asks for a value, retrieve its position instantly and examine the four adjacent or four diagonal directions with boundary checks. Because only a constant number of cells are examined, the query runs in O(1) time.

What is the best approach for Design Neighbor Sum Service?

The most practical solution uses a hash map that stores value → (row, column) coordinates during initialization. Each query performs a constant-time lookup to locate the value and then checks the four adjacent or diagonal positions. This keeps query time O(1) with O(n²) preprocessing space.

Is Design Neighbor Sum Service asked at Google/Amazon/Meta?

Problems involving matrix traversal, hash maps, and lightweight system design patterns are common in interviews at companies like Amazon and Google. While this exact problem may vary, similar tasks that combine grid indexing with constant-time lookups frequently appear in coding rounds.

What data structure is used in Design Neighbor Sum Service?

The core data structure is a hash table mapping each grid value to its coordinates. The grid itself is stored as a 2D array or matrix, and directional offsets are used to check neighboring cells efficiently.

What is the time complexity of Design Neighbor Sum Service?

Initialization requires scanning the entire grid once, which takes O(n²) time. Each query—either adjacentSum or diagonalSum—checks at most four cells, so the runtime per query is O(1). Space complexity is O(n²) due to the value-to-position mapping.

Description Hints 2 Solution Video 5:47

3242. Design Neighbor Sum Service

Easy76.4% acceptance

You are given a n x n 2D array grid containing distinct elements in the range [0, n² - 1].

Implement the NeighborSum class:

NeighborSum(int [][]grid) initializes the object.
int adjacentSum(int value) returns the sum of elements which are adjacent neighbors of value, that is either to the top, left, right, or bottom of value in grid.
int diagonalSum(int value) returns the sum of elements which are diagonal neighbors of value, that is either to the top-left, top-right, bottom-left, or bottom-right of value in grid.

Example 1:

Input:

["NeighborSum", "adjacentSum", "adjacentSum", "diagonalSum", "diagonalSum"]

[[[[0, 1, 2], [3, 4, 5], [6, 7, 8]]], [1], [4], [4], [8]]

Output: [null, 6, 16, 16, 4]

Explanation:

The adjacent neighbors of 1 are 0, 2, and 4.
The adjacent neighbors of 4 are 1, 3, 5, and 7.
The diagonal neighbors of 4 are 0, 2, 6, and 8.
The diagonal neighbor of 8 is 4.

Example 2:

Input:

["NeighborSum", "adjacentSum", "diagonalSum"]

[[[[1, 2, 0, 3], [4, 7, 15, 6], [8, 9, 10, 11], [12, 13, 14, 5]]], [15], [9]]

Output: [null, 23, 45]

Explanation:

The adjacent neighbors of 15 are 0, 10, 7, and 6.
The diagonal neighbors of 9 are 4, 12, 14, and 15.

Constraints:

3 <= n == grid.length == grid[0].length <= 10
0 <= grid[i][j] <= n² - 1
All grid[i][j] are distinct.
value in adjacentSum and diagonalSum will be in the range [0, n² - 1].
At most 2 * n² calls will be made to adjacentSum and diagonalSum.

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Hints open one at a time

Hint 1

Find the cell (i, j) in which the element is present.

Hint 2

You can store the coordinates for each value.

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3242. Design Neighbor Sum Service

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