- mingki 작성
You are given an integer array
heights representing the heights of buildings, some
bricks, and some
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
- 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.
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.
Input: heights = [4,12,2,7,3,18,20,3,19], bricks = 10, ladders = 2 Output: 7
Input: heights = [14,3,19,3], bricks = 17, ladders = 0 Output: 3
1 <= heights.length <= 105
1 <= heights[i] <= 106
0 <= bricks <= 109
0 <= ladders <= heights.length
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: # 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: 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