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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.
This Java solution uses BFS to navigate the shortest path from one tree to another while maintaining the correct order by tree height. The algorithm initializes a list of trees and sorts them by height, then computes the minimal steps between consecutive trees using BFS. If any tree is unreachable, the solution 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.
1from typing import List, Tuple
2from collections import deque
3
4class Solution:
5 directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
6
7 def serialize(self, r: int, c: int, cols: int) -> int:
8 return r * cols + c
9
10 def deserialize(self, key: int, cols: int) -> Tuple[int, int]:
11 return (key // cols, key % cols)
12
13 def bidirectional_bfs(self, forest: List[List[int]], sr: int, sc: int, tr: int, tc: int):
14 if sr == tr and sc == tc:
15 return 0
16 rows, cols = len(forest), len(forest[0])
17 q1 = deque([self.serialize(sr, sc, cols)])
18 q2 = deque([self.serialize(tr, tc, cols)])
19 visited1 = {self.serialize(sr, sc, cols)}
20 visited2 = {self.serialize(tr, tc, cols)}
21 steps = 0
22 while q1 and q2:
23 if len(q1) > len(q2):
24 q1, q2 = q2, q1
25 visited1, visited2 = visited2, visited1
26 steps += 1
27 for _ in range(len(q1)):
28 current = q1.popleft()
29 r, c = self.deserialize(current, cols)
30 for dr, dc in self.directions:
31 nr, nc = r + dr, c + dc
32 if 0 <= nr < rows and 0 <= nc < cols and forest[nr][nc] > 0:
33 nxt = self.serialize(nr, nc, cols)
34 if nxt in visited2:
35 return steps
36 if nxt not in visited1:
37 visited1.add(nxt)
38 q1.append(nxt)
39 return -1
40
41 def cutOffTree(self, forest: List[List[int]]) -> int:
42 trees = [(height, r, c)
43 for r, row in enumerate(forest)
44 for c, height in enumerate(row) if height > 1]
45 trees.sort()
46 sr, sc = 0, 0
47 total_steps = 0
48 for _, tr, tc in trees:
49 steps = self.bidirectional_bfs(forest, sr, sc, tr, tc)
50 if steps == -1:
51 return -1
52 total_steps += steps
53 sr, sc = tr, tc
54 return total_stepsPython realizes the bidirectional search using dual BFS queues for start and target, merging states reached from both ends. It prioritizes advancing the search queue with fewer elements, thereby optimizing space and efficiency. The express serialization of nodes helps manage state transitions seamlessly.