Introduction

In this blog post, we will explore how to solve the LeetCode problem "3242" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

• Precompute Value Positions: Store the row and column indices of each value in the grid using a hash map (dictionary). This allows for O(1) lookup of a value's position.
  • Calculate Sums Based on Position: Given a value, retrieve its row and column from the hash map. Then, calculate the adjacent or diagonal sums by checking the grid cells around that position. Handle edge cases (values at the boundaries of the grid) by checking for valid row and column indices before accessing grid elements.
  • Avoid Redundant Calculations: The sums are calculated directly based on the grid and the precomputed value positions. No dynamic programming or other techniques are necessary as the problem constraints allow for efficient direct computation.

• Runtime Complexity: O(1) for adjacentSum and diagonalSum after the initial O(n^2) initialization. Storage Complexity: O(n^2) for storing value positions in the hash map.

Code

    class NeighborSum:
    def __init__(self, grid: list[list[int]]):
        """
        Initializes the object.
        """
        self.grid = grid
        self.n = len(grid)
        self.value_to_position = {}
        for r in range(self.n):
            for c in range(self.n):
                self.value_to_position[self.grid[r][c]] = (r, c)

    def adjacentSum(self, value: int) -> int:
        """
        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.
        """
        r, c = self.value_to_position[value]
        adjacent_sum = 0
        # Top
        if r > 0:
            adjacent_sum += self.grid[r - 1][c]
        # Bottom
        if r < self.n - 1:
            adjacent_sum += self.grid[r + 1][c]
        # Left
        if c > 0:
            adjacent_sum += self.grid[r][c - 1]
        # Right
        if c < self.n - 1:
            adjacent_sum += self.grid[r][c + 1]

        return adjacent_sum

    def diagonalSum(self, value: int) -> int:
        """
        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.
        """
        r, c = self.value_to_position[value]
        diagonal_sum = 0
        # Top-left
        if r > 0 and c > 0:
            diagonal_sum += self.grid[r - 1][c - 1]
        # Top-right
        if r > 0 and c < self.n - 1:
            diagonal_sum += self.grid[r - 1][c + 1]
        # Bottom-left
        if r < self.n - 1 and c > 0:
            diagonal_sum += self.grid[r + 1][c - 1]
        # Bottom-right
        if r < self.n - 1 and c < self.n - 1:
            diagonal_sum += self.grid[r + 1][c + 1]

        return diagonal_sum