Introduction

In this blog post, we will explore how to solve the LeetCode problem "1631" 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 a hiker preparing for an upcoming hike. You are given heights, a 2D array of size rows x columns, where heights[row][col] represents the height of cell (row, col). You are situated in the top-left cell, (0, 0), and you hope to travel to the bottom-right cell, (rows-1, columns-1) (i.e., 0-indexed). You can move up, down, left, or right, and you wish to find a route that requires the minimum effort. A route's effort is the maximum absolute difference in heights between two consecutive cells of the route. Return the minimum effort required to travel from the top-left cell to the bottom-right cell. Example 1: Input: heights = [[1,2,2],[3,8,2],[5,3,5]] Output: 2 Explanation: The route of [1,3,5,3,5] has a maximum absolute difference of 2 in consecutive cells. This is better than the route of [1,2,2,2,5], where the maximum absolute difference is 3. Example 2: Input: heights = [[1,2,3],[3,8,4],[5,3,5]] Output: 1 Explanation: The route of [1,2,3,4,5] has a maximum absolute difference of 1 in consecutive cells, which is better than route [1,3,5,3,5]. Example 3: Input: heights = [[1,2,1,1,1],[1,2,1,2,1],[1,2,1,2,1],[1,2,1,2,1],[1,1,1,2,1]] Output: 0 Explanation: This route does not require any effort. Constraints: rows == heights.length columns == heights[i].length 1 <= rows, columns <= 100 1 <= heights[i][j] <= 106

Explanation

Here's a solution to the hiking route problem:

• Binary Search on Effort: The minimum effort lies within a range (0 to max_height_difference). Use binary search to find the minimum effort value for which a path exists from the top-left to the bottom-right cell.

• Check Path Existence with BFS/DFS: For a given 'effort' value (determined during binary search), use Breadth-First Search (BFS) or Depth-First Search (DFS) to determine if a path exists between the start and end points. The condition for valid moves is that the absolute height difference between adjacent cells must be less than or equal to the current 'effort'.

• Optimization: BFS is generally preferred here because it can find the shortest path (in terms of number of edges) which will lead to earlier termination if a path exists for a given effort.

• Runtime Complexity: O(rows columns log(max_height_difference)), where max_height_difference is the maximum possible height difference in the input array. Storage Complexity: O(rows * columns) due to the visited array used in BFS.

Code

    from collections import deque

def minimumEffortPath(heights):
    rows, cols = len(heights), len(heights[0])
    left, right = 0, 10**6
    ans = 0

    def is_possible(effort):
        visited = [[False] * cols for _ in range(rows)]
        q = deque([(0, 0)])
        visited[0][0] = True

        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]

        while q:
            row, col = q.popleft()

            if row == rows - 1 and col == cols - 1:
                return True

            for dr, dc in directions:
                new_row, new_col = row + dr, col + dc

                if 0 <= new_row < rows and 0 <= new_col < cols and not visited[new_row][new_col]:
                    if abs(heights[new_row][new_col] - heights[row][col]) <= effort:
                        visited[new_row][new_col] = True
                        q.append((new_row, new_col))
        return False

    while left <= right:
        mid = (left + right) // 2
        if is_possible(mid):
            ans = mid
            right = mid - 1
        else:
            left = mid + 1

    return ans