Cut off trees for golf event [A* Search Algorithm]¶
Time: O(Tx(LogT+MxN)); Space: O(T+MxN); hard
You are asked to cut off trees in a forest for a golf event. The forest is represented as a non-negative 2D map, in this map: 1. 0 represents the obstacle can’t be reached. 2. 1 represents the ground can be walked through. 3. The place with number bigger than 1 represents a tree can be walked through, and this positive number represents the tree’s height.
You are asked to cut off all the trees in this forest in the order of tree’s height - always cut off the tree with lowest height first. And after cutting, the original place has the tree will become a grass (value 1).
You will start from the point (0, 0) and you should output the minimum steps you need to walk to cut off all the trees. If you can’t cut off all the trees, output -1 in that situation.
You are guaranteed 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
Example 2:
Input: forest =
[
[1,2,3],
[0,0,0],
[7,6,5]
]
Output: -1
Example 3:
Input: forest =
[
[2,3,4],
[0,0,5],
[8,7,6]
]
Output: 6
Explanation:
You started from the point (0,0) and you can cut off the tree in (0,0) directly without walking.
Hint: size of the given matrix will not exceed 50x50.
Notes:
1 <= len(forest) <= 50
1 <= len(forest[i]) <= 50
0 <= forest[i][j] <= 10^9
[2]:
import heapq
class Solution1(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 range(m):
for j in range(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
[3]:
s = Solution1()
forest = [
[1,2,3],
[0,0,4],
[7,6,5]
]
assert s.cutOffTree(forest) == 6
forest = [
[1,2,3],
[0,0,0],
[7,6,5]
]
assert s.cutOffTree(forest) == -1
forest = [
[2,3,4],
[0,0,5],
[8,7,6]
]
assert s.cutOffTree(forest) == 6
[4]:
import heapq
import collections
class Solution2(object):
"""
TLE
Time: O(T * (LogT + M * N)), T is the Number of trees
Space: O(T + M * N)
"""
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 range(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 range(m):
for j in range(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
[5]:
s = Solution2()
forest = [
[1,2,3],
[0,0,4],
[7,6,5]
]
assert s.cutOffTree(forest) == 6
forest = [
[1,2,3],
[0,0,0],
[7,6,5]
]
assert s.cutOffTree(forest) == -1
forest = [
[2,3,4],
[0,0,5],
[8,7,6]
]
assert s.cutOffTree(forest) == 6
