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[6/21] 1642. Furthest Building You Can Reach
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Medium
189455Add to ListShareYou are given an integer array heights
representing the heights of buildings, some bricks
, and some ladders
.
You start your journey from building 0
and move to the next building by possibly using bricks or ladders.
While moving from building i
to building i+1
(0-indexed),
- If the current building's height is greater than or equal to the next building's height, you do not need a ladder or bricks.
- If the current building's height is less than the next building's height, you can either use one ladder or
(h[i+1] - h[i])
bricks.
Return the furthest building index (0-indexed) you can reach if you use the given ladders and bricks optimally.
Example 1:
Input: heights = [4,2,7,6,9,14,12], bricks = 5, ladders = 1 Output: 4 Explanation: Starting at building 0, you can follow these steps: - Go to building 1 without using ladders nor bricks since 4 >= 2. - Go to building 2 using 5 bricks. You must use either bricks or ladders because 2 < 7. - Go to building 3 without using ladders nor bricks since 7 >= 6. - Go to building 4 using your only ladder. You must use either bricks or ladders because 6 < 9. It is impossible to go beyond building 4 because you do not have any more bricks or ladders.
Example 2:
Input: heights = [4,12,2,7,3,18,20,3,19], bricks = 10, ladders = 2 Output: 7
Example 3:
Input: heights = [14,3,19,3], bricks = 17, ladders = 0 Output: 3
Constraints:
1 <= heights.length <= 105
1 <= heights[i] <= 106
0 <= bricks <= 109
0 <= ladders <= heights.length
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Runtime: 866 ms, faster than 44.92% of Python3 online submissions for Furthest Building You Can Reach.
Memory Usage: 28.7 MB, less than 16.28% of Python3 online submissions for Furthest Building You Can Reach.
Memory Usage: 28.7 MB, less than 16.28% of Python3 online submissions for Furthest Building You Can Reach.
import collections, heapq
class Solution:
def furthestBuilding(self, heights: List[int], bricks: int, ladders: int) -> int:
reached = 0
max_heap = [] # bricks_used
heapq.heapify(max_heap)
for idx, h in enumerate(heights[:-1]):
if heights[idx+1] <= h:
reached+=1
continue
diff = heights[idx+1] - h
if bricks < diff:
if not ladders: break
# ========= Same as Below if statement======
# if not max_heap or diff > -1*max_heap[0]:
# ladders -= 1
# else:
# bricks += -1*heapq.heappop(max_heap)
# heapq.heappush(max_heap, -1*diff)
# bricks -= diff
# ladders -= 1
# ===============================
#=============Same as Above if-else
if max_heap and diff < -1*max_heap[0]:
bricks += -1*heapq.heappop(max_heap)
heapq.heappush(max_heap, -1*diff)
bricks -= diff
ladders -= 1
# =================================
else:
bricks -= diff
heapq.heappush(max_heap, -1*diff)
reached += 1
return reached