How do you solve Furthest Building You Can Reach on LeetCode?

Furthest Building You Can Reach (LeetCode #1642) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using ladders for the largest jumps is optimal since it preserves bricks for smaller jumps.

What is the brute force solution for Furthest Building You Can Reach?

The brute force approach for Furthest Building You Can Reach is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying every possible combination of using bricks and ladders to see how far we can get. This method is straightforward but inefficient, as it checks each building one by one and calculates the resources needed for each jump. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Furthest Building You Can Reach?

Furthest Building You Can Reach uses the following concepts: Array, Greedy, Heap (Priority Queue). The recommended patterns to study are: Greedy, Heap (Priority Queue).

What are the interview tips for Furthest Building You Can Reach?

Always consider edge cases, such as when you have no bricks or ladders. Be prepared to explain your choice of data structures, like why a min-heap is used. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Furthest Building You Can Reach?

Failing to use ladders for the largest height differences. Not managing the resources carefully, leading to running out of bricks prematurely.

#1642

Furthest Building You Can Reach

Medium

ArrayGreedyHeap (Priority Queue)GreedyHeap (Priority Queue)

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution uses a min-heap (priority queue) to keep track of the jumps we make using bricks. This allows us to efficiently decide when to use a ladder for the largest jump, ensuring we maximize our reach.

⚙️

Algorithm

5 steps

1Step 1: Initialize a min-heap to store the heights of the jumps made with bricks.
2Step 2: For each building, if the next building is taller, calculate the height difference.
3Step 3: Use bricks for the jump and push the height difference into the min-heap.
4Step 4: If the number of jumps exceeds the number of ladders, remove the smallest jump from the heap and add that height to the bricks.
5Step 5: If at any point bricks become negative, return the last building index.

solution.py22 lines

1# Full working Python code
2import heapq
3
4def furthestBuilding(heights, bricks, ladders):
5    min_heap = []
6    n = len(heights)
7    for i in range(n - 1):
8        if heights[i] < heights[i + 1]:
9            diff = heights[i + 1] - heights[i]
10            heapq.heappush(min_heap, diff)
11            bricks -= diff
12            if len(min_heap) > ladders:
13                bricks += heapq.heappop(min_heap)
14            if bricks < 0:
15                return i
16    return n - 1
17
18# Example usage
19heights = [4, 2, 7, 6, 9, 14, 12]
20bricks = 5
21ladders = 1
22print(furthestBuilding(heights, bricks, ladders))

ℹ

Complexity note: The time complexity is O(n log n) because we are using a priority queue (min-heap) to keep track of the jumps, which requires log n time for each insertion and removal.

1Using ladders for the largest jumps is optimal since it preserves bricks for smaller jumps.
2Min-heap allows efficient management of the jumps made with bricks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.