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Cut Off Trees for Golf Event

Number: 675

Difficulty: Hard

Paid? No

Companies: Amazon

Problem Description

Given a forest represented as an m x n matrix where 0 indicates an obstacle, 1 indicates an empty cell, and any number greater than 1 represents a tree (with its height), the goal is to cut off all the trees in order of increasing height. You start at (0, 0) and can move in four directions (up, down, left, right). Each step counts as one move, and when you cut a tree, that cell becomes 1 (an empty cell). If it is impossible to reach all trees, return -1; otherwise, return the minimum number of steps required to cut off all trees.

Key Insights

  • Start by identifying all the trees (cells with value > 1) and sort them by their height.
  • Use a Breadth-First Search (BFS) for finding the shortest path between the current position and the next tree.
  • Update the starting position to the location of the cut tree after each step.
  • If any tree is unreachable (BFS returns -1), the overall solution should return -1.
  • The grid size is limited to 50 x 50, making BFS operations efficient despite repeated searches.

Space and Time Complexity

Time Complexity: O(T * m * n) where T is the number of trees. In the worst case, each BFS may scan the entire grid. Space Complexity: O(m * n) for storing the visitation states during BFS.

Solution

The approach first involves collecting and sorting all trees by height. For each tree, we use BFS to compute the minimum steps needed from the current position to that tree. We maintain a queue for BFS and mark cells as visited to avoid cycles. After reaching a tree, we update our start position and add the number of steps taken to a running total. If at any point the BFS is unable to reach the desired tree, we return -1. This procedure continues until all trees are processed. The use of a priority queue is not necessary because we only require sorting by tree height once at the beginning.

Code Solutions

from collections import deque

def cutOffTree(forest):
    # Get rows and columns
    rows, cols = len(forest), len(forest[0])
    
    # Calculate positions of trees and sort them by height.
    trees = sorted((height, i, j) for i in range(rows) for j in range(cols) if forest[i][j] > 1)
    
    def bfs(start_i, start_j, target_i, target_j):
        # Perform BFS to find the minimum steps from (start_i, start_j) to (target_i, target_j)
        if start_i == target_i and start_j == target_j:
            return 0
        visited = [[False] * cols for _ in range(rows)]
        q = deque([(start_i, start_j, 0)])
        visited[start_i][start_j] = True
        # Four possible movements: up, down, left, right.
        directions = [(1,0), (-1,0), (0,1), (0,-1)]
        
        while q:
            i, j, steps = q.popleft()
            for di, dj in directions:
                ni, nj = i + di, j + dj
                # Check if we are within bounds and not visited and not an obstacle.
                if 0 <= ni < rows and 0 <= nj < cols and not visited[ni][nj] and forest[ni][nj] != 0:
                    if ni == target_i and nj == target_j:
                        return steps + 1
                    visited[ni][nj] = True
                    q.append((ni, nj, steps + 1))
        return -1

    total_steps = 0
    start_i = start_j = 0
    # Process each tree in the sorted order.
    for _, tree_i, tree_j in trees:
        steps = bfs(start_i, start_j, tree_i, tree_j)
        if steps == -1:
            return -1
        total_steps += steps
        start_i, start_j = tree_i, tree_j

    return total_steps

# Test Example
forest = [[1,2,3],[0,0,4],[7,6,5]]
print(cutOffTree(forest))  # Expected output: 6
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