Problem Description
Given an array of building heights, along with a fixed number of bricks and ladders, determine the furthest building index you can reach by optimally deciding whether to use bricks or ladders when moving from a lower building to a taller one. When moving between adjacent buildings, you must cover the difference in heights with either ladders (which can cover any height) or bricks (which cover exactly the height difference), unless the next building is of equal or lesser height.
Key Insights
- Use a greedy approach to prioritize ladders for the largest jumps.
- Use bricks for smaller height differences.
- Maintain a min-heap (priority queue) to track the height differences where you chose to use bricks initially.
- When the number of climbs exceeds the available ladders, the smallest jump (from the heap) is replaced by using bricks.
- The simulation stops when the bricks run out before you can cover the required jump.
Space and Time Complexity
Time Complexity: O(n log n) in the worst-case scenario due to the heap operations, where n is the number of buildings. Space Complexity: O(n) for the worst-case heap storage.
Solution
We solve the problem by simulating the journey from the first building to the last. For each jump, if the next building is taller than the current one, we calculate the height difference. We push this difference into a min-heap. If the number of required climbs (i.e., the heap's size) exceeds the available ladders, we remove the smallest jump from the heap and use bricks to cover it instead. If at any point the bricks become insufficient (i.e., bricks drop below zero), we cannot progress further and return the current building index. This approach optimally allocates ladders to the largest jumps while using bricks for the smaller ones.