Introduction

In this blog post, we will explore how to solve the LeetCode problem "1642" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 Constraints: 1 <= heights.length <= 105 1 <= heights[i] <= 106 0 <= bricks <= 109 0 <= ladders <= heights.length

Explanation

Here's the solution to the problem, focusing on efficiency and optimality:

• Key Idea: Use a min-heap (priority queue) to store the differences in height that we have used bricks for so far. This allows us to dynamically decide, at each building, whether to use bricks for the current climb, or to swap bricks for a ladder on one of the previous climbs that used the least amount of bricks.

• Greedy Ladder Assignment: The core idea is to always use ladders for the largest height differences encountered so far. The heap helps to keep track of the smallest height differences for brick usage. If at any point we run out of bricks, we attempt to replace a brick usage with a ladder usage. If even after doing so, we run out of bricks, we cannot proceed further.

• Runtime & Space Complexity: O(N log L) time, where N is the number of buildings and L is the number of ladders. The space complexity is O(L) due to the heap.

Code

    import heapq

def furthestBuilding(heights, bricks, ladders):
    """
    Finds the furthest building index that can be reached with the given bricks and ladders.

    Args:
        heights: A list of integers representing the heights of buildings.
        bricks: The number of bricks available.
        ladders: The number of ladders available.

    Returns:
        The furthest building index that can be reached.
    """

    n = len(heights)
    height_diffs = []  # Min-heap to store height differences we've used bricks for

    for i in range(n - 1):
        diff = heights[i + 1] - heights[i]
        if diff <= 0:
            continue  # No need to use bricks or ladders

        heapq.heappush(height_diffs, diff)  # Try to use bricks first
        bricks -= diff

        if bricks < 0:  # If we've run out of bricks
            if ladders > 0:
                smallest_bricks = heapq.heappop(height_diffs)  # Replace smallest brick usage with a ladder
                bricks += smallest_bricks
                ladders -= 1
            else:
                return i  # Cannot proceed further

    return n - 1  # Reached the last building