Day88_Furthest Building You Can Reach.py

# Furthest Building You Can Reach                        Difficulty = 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


class Solution:
    def furthestBuilding(self, heights: List[int], bricks: int, ladders: int) -> int:
        min_heap = []
        n = len(heights)
        
        for i in range(n-1):
            climb = heights[i+1] - heights[i]
            
            if climb <= 0:
                continue
            
            # we need to use a ladder or some bricks, always take the ladder at first
            if climb > 0:
                heapq.heappush(min_heap, climb)
            
            # ladders are all in used, find the current shortest climb to use bricks instead!
            if len(min_heap) > ladders:
                # find the current shortest climb to use bricks
                brick_need = heapq.heappop(min_heap)
                bricks -= brick_need
            
            if bricks < 0:
                return i
        
        return n-1