# 1642 Furthest Building You Can Reach

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:

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# #
# #
/ # # #
/ # # #
- # / # # #
- # # # # #
- # # # # #
# - # # # # #
# - # # # # #
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# # # # # # #

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.
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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`
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 `````` ``````class Solution: def furthestBuilding(self, heights: List[int], bricks: int, ladders: int) -> int: n = len(heights) # use ladders for the highest climbs heap = [] for i in range(1, n): diff = heights[i] - heights[i - 1] if diff <= 0: continue heapq.heappush(heap, diff) # If haven't gone over the num of ladders, nothing else to do if len(heap) <= ladders: continue # Otherwise, we have to take a climb out of heap bricks -= heapq.heappop(heap) if bricks < 0: return i - 1 return n - 1``````