Furthest Building You Can Reach

lcpy gen -n 1642   # by problem number
lcpy gen -s furthest_building_you_can_reach   # by problem name

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

import heapq


class Solution:
    # Time: O(n log k) where k = ladders
    # Space: O(k)
    def furthest_building(self, heights: list[int], bricks: int, ladders: int) -> int:
        climbs: list[int] = []
        for i in range(len(heights) - 1):
            climb = heights[i + 1] - heights[i]
            if climb <= 0:
                continue
            heapq.heappush(climbs, climb)
            if len(climbs) > ladders:
                bricks -= heapq.heappop(climbs)
            if bricks < 0:
                return i
        return len(heights) - 1