1642 Furthest Building You Can Reach

🔀 Pick One

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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  - # # # # #
# - # # # # #
# - # # # # #
# # # # # # #
# # # # # # #

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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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