3242. Design Neighbor Sum Service

Problem

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.

https://leetcode.cn/problems/design-neighbor-sum-service/

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:

case1

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:

case2

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.

Test Cases

    python
    
class NeighborSum:

    def __init__(self, grid: List[List[int]]):

    def adjacentSum(self, value: int) -> int:

    def diagonalSum(self, value: int) -> int:

# Your NeighborSum object will be instantiated and called as such:
# obj = NeighborSum(grid)
# param_1 = obj.adjacentSum(value)
# param_2 = obj.diagonalSum(value)

solution_test.py下载完整源码

    py
    
  
import pytest

from solution import NeighborSum


@pytest.mark.parametrize('actions, params, expects', [
    (
        ["NeighborSum", "adjacentSum", "adjacentSum", "diagonalSum", "diagonalSum"],
        [[[[0, 1, 2], [3, 4, 5], [6, 7, 8]]], [1], [4], [4], [8]],
        [None, 6, 16, 16, 4],
    ),
    (
        ["NeighborSum", "adjacentSum", "diagonalSum"],
        [[[[1, 2, 0, 3], [4, 7, 15, 6], [8, 9, 10, 11], [12, 13, 14, 5]]], [15], [9]],
        [None, 23, 45],
    ),
])
def test_neighbor_sum(actions, params, expects):
    ns = None
    for action, param, expected in zip(actions, params, expects):
        if action == 'NeighborSum':
            ns = NeighborSum(*param)
        elif action == 'adjacentSum':
            assert ns.adjacentSum(*param) == expected
        elif action == 'diagonalSum':
            assert ns.diagonalSum(*param) == expected

Thoughts

在初始化的时候直接把所有格子的 adjacentSum 和 diagonalSum 计算好保存下来。

因为元素的值刚好是 0 到 n² - 1，可以直接用作数组，元素值即为数字下标。

每次查询的时间复杂度为 O(1)。

构造的时间复杂度为 O(n²)，空间复杂度亦然。

构造的计算有两个方向，一是遍历到一个格子时，计算它的 adjacent 格子和 diagonal 格子的数字之和。另一是遍历到一个格子时，计算它对周围格子的 adjacentSum 和 diagonalSum 的贡献。

注意矩阵的边界。

Code

solution.py下载完整源码

    py
    
  
from typing import List


class NeighborSum:

    def __init__(self, grid: List[List[int]]):
        n = len(grid)
        n2 = n * n
        self._adjacent_sums = [0] * n2
        self._diagonal_sums = [0] * n2

        g = lambda x, y: grid[x][y] if 0 <= x < n and 0 <= y < n else 0

        for i in range(n):
            for j in range(n):
                v = grid[i][j]
                self._adjacent_sums[v] = g(i-1, j) + g(i, j-1) + g(i, j+1) + g(i+1, j)
                self._diagonal_sums[v] = g(i-1, j-1) + g(i-1, j+1) + g(i+1, j-1) + g(i+1, j+1)

    def adjacentSum(self, value: int) -> int:
        return self._adjacent_sums[value]

    def diagonalSum(self, value: int) -> int:
        return self._diagonal_sums[value]


# Your NeighborSum object will be instantiated and called as such:
# obj = NeighborSum(grid)
# param_1 = obj.adjacentSum(value)
# param_2 = obj.diagonalSum(value)

#Problem

#Test Cases

#Thoughts

#Code

Problem

Test Cases

Thoughts

Code