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# [6/21] 1642. Furthest Building You Can Reach

• mingki 작성
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• 65 조회
• 1 댓글

### 본문

Medium

You 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`

댓글 1

## 학부유학생님의 댓글

• 학부유학생
• 작성일
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.
``````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:

# ========= Same as Below if statement======
# if not max_heap or diff > -1*max_heap[0]:
# else:
#     bricks += -1*heapq.heappop(max_heap)
#     heapq.heappush(max_heap, -1*diff)
#     bricks -= diff
# ===============================
#=============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
# =================================

else:
bricks -= diff
heapq.heappush(max_heap, -1*diff)
reached += 1

return reached

``````
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