We use a Min-Heap to keep track of the largest height differences where we used a ladder. If the number of height differences exceeds the number of ladders, we replace the smallest difference (using the ladder) with bricks.

1. Initialize a min-heap h.
2. For each building from i = 0 to len(heights) - 2:
   • Calculate diff = heights[i+1] - heights[i].
   • If diff > 0:
     • Push diff to the heap.
     • If len(h) > ladders:
       • Pop the smallest diff from the heap.
       • Subtract that diff from bricks.
     • If bricks < 0, return current index i.
3. If we finish the loop, return len(heights) - 1.

• Time Complexity: O(N log L), where L is the number of ladders.
• Space Complexity: O(L).

import heapq

def furthest_building(heights, bricks, ladders):
    heap = []
    
    for i in range(len(heights) - 1):
        diff = heights[i+1] - heights[i]
        if diff > 0:
            heapq.heappush(heap, diff)
            if len(heap) > ladders:
                bricks -= heapq.heappop(heap)
            if bricks < 0:
                return i
                
    return len(heights) - 1