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.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.lengthn == forest[i].length1 <= m, n <= 500 <= forest[i][j] <= 109In #675 Cut Off Trees for Golf Event, the forest is represented as a matrix where each cell may contain a tree with a certain height. The objective is to cut all trees in increasing order of height while minimizing the total steps taken. A common strategy is to first collect all trees and sort them by height (often using a list or priority queue). Starting from position (0,0), you then repeatedly move to the next tree in sorted order.
To compute the shortest path between the current position and the next tree, apply Breadth-First Search (BFS) on the grid. BFS works well because each move has equal cost and the grid may contain obstacles (cells with value 0). If any tree cannot be reached, the result is -1. This approach essentially performs multiple shortest-path searches across the matrix while processing trees in sorted order.
The overall efficiency depends on the number of trees and repeated BFS traversals across the grid.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Sort Trees + BFS for Shortest Path | O(T * m * n) | O(m * n) |
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This approach involves using a BFS algorithm to find the shortest path between the starting point and each tree in increasing order of their heights. We first list all trees with their positions and heights, sort them, and for each sorted tree, compute the shortest path using BFS. If any tree is unreachable, we return -1.
The time complexity of this approach is O(T * m * n) where T is the number of trees. This is because we perform BFS from the starting position to each tree. The space complexity is O(m * n) due to the storage of the queue and visited matrix in BFS.
1#include <stdio.h>
2#include <stdlib.h>
3#include <stdbool.h>
4#include <limits.h>
5
6// Define directions for moving in the grid
This C solution first uses BFS to calculate the shortest path between trees, starting from (0, 0) and moving to trees in increasing order of their height. The binary search is used to sort the trees based on their height first. If any tree is unreachable during a BFS computation, the function returns -1.
Using Bidirectional BFS can help in optimizing the pathfinding step by searching from both the start and the goal concurrently. The aim is to decrease the search space substantially as the two searches meet in the middle. This method is an extension to the regular BFS approach, potentially optimizing the traversal cost considerably under specific configurations.
The potential benefits of bidirectional search in space complexity can be realized when effectively reducing the number of explored nodes, as time complexity ideally shortens from O(b^d) to roughly O(b^(d/2)), where b is the branching factor and d the solution depth.
1Implementing a complete bidirectional BFS in C is complex and often involves intricate management of forward and backward search queues and visited states. This might require substantial utility function support forWatch expert explanations and walkthroughs
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BFS guarantees the shortest path in an unweighted grid where each move has equal cost. Since the player can move in four directions and obstacles exist, BFS helps determine the minimum number of steps between trees.
Yes, this problem is considered a challenging grid traversal question and can appear in interviews at companies like Google, Amazon, or Meta. It tests BFS, matrix traversal, and the ability to combine sorting with repeated shortest-path searches.
A list or priority queue is useful for sorting trees by height, while a queue is used for BFS traversal of the grid. BFS is ideal because it finds the shortest path in an unweighted grid efficiently.
The optimal approach is to first collect all trees and sort them by height. Then repeatedly run BFS from the current position to the next tree in sorted order to find the shortest path. If any tree is unreachable, return -1.
A detailed explanation of handling bidirectional BFS in C would involve structuring two queues for the forward and backward search, ensuring efficient state management overlaps and transitions by interfacing node positions from both directions dynamically, essentially shortening the BFS travel path.