Leetcode #1642: Furthest Building You Can Reach
In this guide, we solve Leetcode #1642 Furthest Building You Can Reach in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, Array, Heap (Priority Queue)
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
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.
Python Solution
class Solution:
def furthestBuilding(self, heights: List[int], bricks: int, ladders: int) -> int:
h = []
for i, a in enumerate(heights[:-1]):
b = heights[i + 1]
d = b - a
if d > 0:
heappush(h, d)
if len(h) > ladders:
bricks -= heappop(h)
if bricks < 0:
return i
return len(heights) - 1
Complexity
The time complexity is O(n log n). The space complexity is O(1) to O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.