How do you solve Cut Off Trees for Golf Event on LeetCode?

Cut Off Trees for Golf Event (LeetCode #675) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n + n²) time complexity and O(n) space complexity. Key insight: Using BFS ensures that we explore all possible paths efficiently.

What is the time complexity of Cut Off Trees for Golf Event?

The optimal solution for Cut Off Trees for Golf Event runs in O(n log n + n²) time and uses O(n) space. The time complexity is O(n log n) for sorting the trees and O(n²) for BFS in the worst case, leading to an overall complexity of O(n log n + n²).

What is the brute force solution for Cut Off Trees for Golf Event?

The brute force approach for Cut Off Trees for Golf Event is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves finding the shortest path to each tree in the order of their heights. We can use a breadth-first search (BFS) to explore all possible paths from the starting point to each tree, ensuring we find the minimum steps required. The optimal Optimal Solution approach improves this to O(n log n + n²).

What algorithm or data structure is used to solve Cut Off Trees for Golf Event?

Cut Off Trees for Golf Event uses the following concepts: Array, Breadth-First Search, Heap (Priority Queue), Matrix. The recommended patterns to study are: .

What are common mistakes when solving Cut Off Trees for Golf Event?

Not considering blocked paths which can lead to unreachable trees.

#675

Cut Off Trees for Golf Event

Hard

ArrayBreadth-First SearchHeap (Priority Queue)Matrix

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n + n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n + n²)Space O(n)

The optimal solution leverages a priority queue to always process the nearest tree first, ensuring that we minimize the number of steps taken. This approach efficiently finds the shortest paths to each tree in the required order.

⚙️

Algorithm

3 steps

1Step 1: Extract and sort the trees by height, storing their coordinates.
2Step 2: Use a priority queue to perform a BFS from the current position to the next tree's position, ensuring we always take the shortest path.
3Step 3: Sum the steps for all trees. If any tree is unreachable, return -1.

solution.py35 lines

1from collections import deque
2import heapq
3
4def cutOffTree(forest):
5    def bfs(start, end):
6        if start == end:
7            return 0
8        queue = deque([start])
9        visited = set([start])
10        steps = 0
11        directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
12        while queue:
13            for _ in range(len(queue)):
14                x, y = queue.popleft()
15                for dx, dy in directions:
16                    nx, ny = x + dx, y + dy
17                    if (0 <= nx < len(forest) and 0 <= ny < len(forest[0]) and
18                        (nx, ny) not in visited and forest[nx][ny] > 0):
19                        if (nx, ny) == end:
20                            return steps + 1
21                        visited.add((nx, ny))
22                        queue.append((nx, ny))
23            steps += 1
24        return -1
25
26    trees = sorted([(forest[i][j], (i, j)) for i in range(len(forest)) for j in range(len(forest[0])) if forest[i][j] > 1])
27    total_steps = 0
28    start = (0, 0)
29    for height, (x, y) in trees:
30        steps = bfs(start, (x, y))
31        if steps == -1:
32            return -1
33        total_steps += steps
34        start = (x, y)
35    return total_steps

ℹ

Complexity note: The time complexity is O(n log n) for sorting the trees and O(n²) for BFS in the worst case, leading to an overall complexity of O(n log n + n²).

1Using BFS ensures that we explore all possible paths efficiently.
2Sorting the trees by height allows us to prioritize cutting the shortest trees first.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.