Watch 10 video solutions for Design Neighbor Sum Service, a easy level problem involving Array, Hash Table, Design. This walkthrough by stoney codes has 750,617 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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:
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:
Constraints:
3 <= n == grid.length == grid[0].length <= 100 <= grid[i][j] <= n2 - 1grid[i][j] are distinct.value in adjacentSum and diagonalSum will be in the range [0, n2 - 1].2 * n2 calls will be made to adjacentSum and diagonalSum.