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Problem solution · Python

Cut Off Trees for Golf Event

Cut Off Trees for Golf Event: a Python solution using breadth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Breadth-first search
Source
Kamyu LeetCode Solutions
Length
102 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Breadth-first search

For Cut Off Trees for Golf Event, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 102 lines of Python from the credited upstream file cut-off-trees-for-golf-event.py.
  • The implementation visibly relies on sequence storage, ordered lookup, work queue.
  • No explicit loop blocks detected.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCut Off Trees for Golf Event · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(t * (logt + m * n)), t is the number of trees# Space: O(t + m * n) import collectionsimport heapq  class Solution(object):    def cutOffTree(self, forest):        """        :type forest: List[List[int]]        :rtype: int        """        def dot(p1, p2):            return p1[0]*p2[0]+p1[1]*p2[1]         def minStep(p1, p2):            min_steps = abs(p1[0]-p2[0])+abs(p1[1]-p2[1])            closer, detour = [p1], []            lookup = set()            while True:                if not closer:  # cannot find a path in the closer expansions                    if not detour:  # no other possible path                        return -1                    # try other possible paths in detour expansions with extra 2-step cost                    min_steps += 2                    closer, detour = detour, closer                i, j = closer.pop()                if (i, j) == p2:                    return min_steps                if (i, j) not in lookup:                    lookup.add((i, j))                    for I, J in (i+1, j), (i-1, j), (i, j+1), (i, j-1):                        if 0 <= I < m and 0 <= J < n and forest[I][J] and (I, J) not in lookup:                            is_closer = dot((I-i, J-j), (p2[0]-i, p2[1]-j)) > 0                            (closer if is_closer else detour).append((I, J))            return min_steps         m, n = len(forest), len(forest[0])        min_heap = []        for i in xrange(m):            for j in xrange(n):                if forest[i][j] > 1:                    heapq.heappush(min_heap, (forest[i][j], (i, j)))         start = (0, 0)        result = 0        while min_heap:            tree = heapq.heappop(min_heap)            step = minStep(start, tree[1])            if step < 0:                return -1            result += step            start = tree[1]        return result  # Time:  O(t * (logt + m * n)), t is the number of trees# Space: O(t + m * n)class Solution_TLE(object):    def cutOffTree(self, forest):        """        :type forest: List[List[int]]        :rtype: int        """        def minStep(p1, p2):            min_steps = 0            lookup = {p1}            q = collections.deque([p1])            while q:                size = len(q)                for _ in xrange(size):                    (i, j) = q.popleft()                    if (i, j) == p2:                        return min_steps                    for i, j in (i+1, j), (i-1, j), (i, j+1), (i, j-1):                        if not (0 <= i < m and 0 <= j < n and forest[i][j] and (i, j) not in lookup):                            continue                        q.append((i, j))                        lookup.add((i, j))                min_steps += 1            return -1         m, n = len(forest), len(forest[0])        min_heap = []        for i in xrange(m):            for j in xrange(n):                if forest[i][j] > 1:                    heapq.heappush(min_heap, (forest[i][j], (i, j)))         start = (0, 0)        result = 0        while min_heap:            tree = heapq.heappop(min_heap)            step = minStep(start, tree[1])            if step < 0:                return -1            result += step            start = tree[1]        return result

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