Introduction

In this blog post, we will explore how to solve the LeetCode problem "675" 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 asked to cut off all the trees in a forest for a golf event. The forest is represented as an m x n matrix. In this matrix: 0 means the cell cannot be walked through. 1 represents an empty cell that can be walked through. A number greater than 1 represents a tree in a cell that can be walked through, and this number is the tree's height. In one step, you can walk in any of the four directions: north, east, south, and west. If you are standing in a cell with a tree, you can choose whether to cut it off. You must cut off the trees in order from shortest to tallest. When you cut off a tree, the value at its cell becomes 1 (an empty cell). Starting from the point (0, 0), return the minimum steps you need to walk to cut off all the trees. If you cannot cut off all the trees, return -1. Note: The input is generated such that no two trees have the same height, and there is at least one tree needs to be cut off. Example 1: Input: forest = [[1,2,3],[0,0,4],[7,6,5]] Output: 6 Explanation: Following the path above allows you to cut off the trees from shortest to tallest in 6 steps. Example 2: Input: forest = [[1,2,3],[0,0,0],[7,6,5]] Output: -1 Explanation: The trees in the bottom row cannot be accessed as the middle row is blocked. Example 3: Input: forest = [[2,3,4],[0,0,5],[8,7,6]] Output: 6 Explanation: You can follow the same path as Example 1 to cut off all the trees. Note that you can cut off the first tree at (0, 0) before making any steps. Constraints: m == forest.length n == forest[i].length 1 <= m, n <= 50 0 <= forest[i][j] <= 109 Heights of all trees are distinct.

Explanation

Here's the solution:

• Identify and Sort Trees: Extract the locations and heights of all trees (values > 1) in the forest. Sort these trees based on their height in ascending order.
• A* Search for Each Tree: Starting from (0, 0), use A* search (or Dijkstra's) to find the shortest path to each tree in the sorted list. After cutting a tree, update the forest by setting the cell value to 1.

• Accumulate Steps: Sum the steps taken to reach each tree. If any tree is unreachable, return -1.

• Runtime Complexity: O(m*n*t*log(m*n)) where t is the number of trees and m, n are the dimensions of the forest, Storage Complexity: O(m*n)

Code

    import heapq

def cutOffTree(forest):
    trees = []
    for r in range(len(forest)):
        for c in range(len(forest[0])):
            if forest[r][c] > 1:
                trees.append((forest[r][c], r, c))

    trees.sort()

    start_row, start_col = 0, 0
    total_steps = 0

    for height, target_row, target_col in trees:
        steps = astar(forest, start_row, start_col, target_row, target_col)
        if steps == -1:
            return -1
        total_steps += steps
        start_row, start_col = target_row, target_col
        forest[target_row][target_col] = 1  # Cut the tree

    return total_steps

def astar(forest, start_row, start_col, target_row, target_col):
    rows = len(forest)
    cols = len(forest[0])

    queue = [(0, 0, start_row, start_col)]  # (f_score, g_score, row, col)
    visited = set()

    while queue:
        f_score, g_score, row, col = heapq.heappop(queue)

        if (row, col) == (target_row, target_col):
            return g_score

        if (row, col) in visited:
            continue
        visited.add((row, col))

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

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

            if 0 <= new_row < rows and 0 <= new_col < cols and forest[new_row][new_col] != 0:
                new_g_score = g_score + 1
                heuristic = abs(new_row - target_row) + abs(new_col - target_col)
                new_f_score = new_g_score + heuristic
                heapq.heappush(queue, (new_f_score, new_g_score, new_row, new_col))

    return -1