How do you solve Design Neighbor Sum Service on LeetCode?

Design Neighbor Sum Service (LeetCode #3242) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a hash map for coordinate storage allows for O(1) access time for neighbor lookups.

What is the brute force solution for Design Neighbor Sum Service?

The brute force approach for Design Neighbor Sum Service is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach involves scanning the entire grid to find the position of the given value and then checking its neighbors. It's straightforward but inefficient for larger grids. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Design Neighbor Sum Service?

Design Neighbor Sum Service uses the following concepts: Array, Hash Table, Design, Matrix, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Design Neighbor Sum Service?

Clearly explain your thought process and the steps you are taking to solve the problem. Consider edge cases and how your solution handles them. Practice writing clean and efficient code, as readability is important in interviews.

What are common mistakes when solving Design Neighbor Sum Service?

Failing to check boundary conditions when accessing neighbors, which can lead to index out of bounds errors. Not using a hash map to store positions, resulting in inefficient searches.

#3242

Design Neighbor Sum Service

Easy

ArrayHash TableDesignMatrixSimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

By storing the positions of each value in a hash map, we can quickly access the coordinates of any value, allowing us to compute sums in constant time.

⚙️

Algorithm

3 steps

1Step 1: Create a hash map to store the coordinates of each value in the grid during initialization.
2Step 2: For adjacentSum, retrieve the coordinates of the value and sum the values of its four adjacent neighbors using the hash map.
3Step 3: For diagonalSum, retrieve the coordinates and sum the values of its four diagonal neighbors using the hash map.

solution.py23 lines

1class NeighborSum:
2    def __init__(self, grid):
3        self.grid = grid
4        self.n = len(grid)
5        self.position_map = {grid[i][j]: (i, j) for i in range(self.n) for j in range(self.n)}
6
7    def adjacentSum(self, value):
8        i, j = self.position_map[value]
9        sum_adj = 0
10        if i > 0: sum_adj += self.grid[i-1][j]  # up
11        if i < self.n - 1: sum_adj += self.grid[i+1][j]  # down
12        if j > 0: sum_adj += self.grid[i][j-1]  # left
13        if j < self.n - 1: sum_adj += self.grid[i][j+1]  # right
14        return sum_adj
15
16    def diagonalSum(self, value):
17        i, j = self.position_map[value]
18        sum_diag = 0
19        if i > 0 and j > 0: sum_diag += self.grid[i-1][j-1]  # top-left
20        if i > 0 and j < self.n - 1: sum_diag += self.grid[i-1][j+1]  # top-right
21        if i < self.n - 1 and j > 0: sum_diag += self.grid[i+1][j-1]  # bottom-left
22        if i < self.n - 1 and j < self.n - 1: sum_diag += self.grid[i+1][j+1]  # bottom-right
23        return sum_diag

ℹ

Complexity note: The time complexity is O(n) for the lookup of coordinates, as we use a hash map. The space complexity is O(n) due to storing the positions of each element.

1Using a hash map for coordinate storage allows for O(1) access time for neighbor lookups.
2Understanding the difference between adjacent and diagonal neighbors is crucial for implementing the sums correctly.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.