Problem Description
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. Return the furthest building index (0-indexed) you can reach if you use the given ladders and bricks optimally.
Key Insights
- You can move to the next building without using any resources if the current building's height is greater than or equal to the next building's height.
- If the current building's height is less than the next, you can either use a ladder or a specific number of bricks equivalent to the height difference.
- The goal is to use ladders for the largest height differences to maximize the distance you can travel.
Space and Time Complexity
Time Complexity: O(n log n)
Space Complexity: O(n)
Solution
To solve this problem, we can implement a greedy approach using a min-heap (priority queue) to keep track of the height differences when we decide to use bricks. As we traverse the heights array, we will check if the next building is taller. If it is, we calculate the height difference and decide whether to use a ladder or bricks. We will always prefer to use ladders for the largest height differences. If we run out of bricks, we will use the smallest height difference recorded in the heap (which corresponds to a previously used brick) to replace it with a ladder.