3242 - Design Neighbor Sum Service.

Easy

You are given a n x n 2D array grid containing distinct elements in the range <code>0, n² - 1</code>.

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

• <code>0 <= gridj<= n² - 1</code>

• All grid[i][j] are distinct.

• value in adjacentSum and diagonalSum will be in the range <code>0, n² - 1</code>.

• At most <code>2 * n²</code> calls will be made to adjacentSum and diagonalSum.