# 1642. Furthest Building You Can Reach import heapq """ 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 height of next building, 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 """ class Solution(object): def furthestBuilding(self, heights, bricks, ladders): heap = [] for i in range(len(heights) - 1): d = heights[i + 1] - heights[i] if d > 0: heapq.heappush(heap, d) if len(heap) > ladders: bricks -= heapq.heappop(heap) if bricks < 0: return i return len(heights) - 1 if __name__ == "__main__": obj = Solution() print(obj.furthestBuilding([4, 2, 7, 6, 9, 14, 12], 5, 1)) # expect: 4 print(obj.furthestBuilding([4, 12, 2, 7, 3, 18, 20, 3, 19], 10, 2)) # expect: 7 print(obj.furthestBuilding([14, 3, 19, 3], 17, 0)) # expect: 3