Solution Explanation

This problem can be efficiently solved using a greedy approach combined with a min-heap (priority queue). The core idea is to prioritize using bricks for smaller height differences and ladders for larger ones. This ensures we maximize our reach given the limited resources.

Algorithm:

1. Initialization: We maintain a min-heap h to store the height differences encountered so far that require either bricks or ladders.

2. Iteration: We iterate through the heights array, comparing each building's height with the next one.

3. Height Difference: If the next building is taller (d > 0), we add the height difference d to the heap.

4. Ladder/Brick Decision: If the heap size exceeds the number of available ladders (len(h) > ladders), it means we need to optimize brick usage. We remove the smallest height difference from the heap (using heappop) and deduct it from our bricks. If after this deduction bricks becomes negative, it implies that we've run out of resources, so we return the current index i.

5. Reaching the End: If the loop completes without running out of bricks or ladders, it means we can reach the last building, so we return len(heights) - 1.

Time Complexity Analysis:

• The main loop iterates through the heights array once, taking O(N) time, where N is the length of the array.
• Heap operations (insertion and deletion) take O(log K) time in the worst case, where K is the number of ladders (at most the number of height differences we need to climb). Since K <= N, the overall time complexity of heap operations is O(N log N).
• Therefore, the dominant factor is O(N log N).

Space Complexity Analysis:

The space complexity is O(K) due to the heap, where K is the number of ladders (or the maximum number of height differences considered). In the worst case, K could be O(N), resulting in O(N) space complexity.

Code Explanation (Python):

import heapq
 
class Solution:
    def furthestBuilding(self, heights: List[int], bricks: int, ladders: int) -> int:
        h = []  # Min-heap to store height differences
        for i, a in enumerate(heights[:-1]):
            b = heights[i + 1]
            d = b - a
            if d > 0:
                heapq.heappush(h, d) # Add height difference to heap
                if len(h) > ladders: #Check if more differences than ladders
                    bricks -= heapq.heappop(h) #Use bricks for smallest difference
                    if bricks < 0:
                        return i #Ran out of resources
        return len(heights) - 1 #Reached end

The other language implementations follow the same logic, using their respective priority queue implementations (PriorityQueue in Java, priority_queue in C++, and a custom heap in Go). The core algorithm remains consistent across all the provided solutions.