Leetcode #675: Cut Off Trees for Golf Event
In this guide, we solve Leetcode #675 Cut Off Trees for Golf Event in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Breadth-First Search, Array, Matrix, Heap (Priority Queue)
Intuition
We need to repeatedly access the smallest or largest element as the input changes.
A heap provides fast insertions and removals while keeping order.
Approach
Push candidates into the heap as you scan, and pop when you need the best element.
Keep the heap size bounded if the problem requires a top-k structure.
Steps:
- Push candidates into a heap.
- Pop the best candidate when needed.
- Maintain heap size or invariants.
Example
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.
Python Solution
class Solution:
def cutOffTree(self, forest: List[List[int]]) -> int:
def f(i, j, x, y):
return abs(i - x) + abs(j - y)
def bfs(i, j, x, y):
q = [(f(i, j, x, y), i, j)]
dist = {i * n + j: 0}
while q:
_, i, j = heappop(q)
step = dist[i * n + j]
if (i, j) == (x, y):
return step
for a, b in [[0, -1], [0, 1], [-1, 0], [1, 0]]:
c, d = i + a, j + b
if 0 <= c < m and 0 <= d < n and forest[c][d] > 0:
if c * n + d not in dist or dist[c * n + d] > step + 1:
dist[c * n + d] = step + 1
heappush(q, (dist[c * n + d] + f(c, d, x, y), c, d))
return -1
m, n = len(forest), len(forest[0])
trees = [
(forest[i][j], i, j) for i in range(m) for j in range(n) if forest[i][j] > 1
]
trees.sort()
i = j = 0
ans = 0
for _, x, y in trees:
t = bfs(i, j, x, y)
if t == -1:
return -1
ans += t
i, j = x, y
return ans
Complexity
The time complexity is O(n log n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.