What is the time complexity of Furthest Building You Can Reach?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Furthest Building You Can Reach?

This problem uses Array, Greedy, Heap (Priority Queue). Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Furthest Building You Can Reach?

Insight 1: The problem is fundamentally about resource allocation and optimization. We need to decide when to use bricks and when to use ladders to maximize our progress. Insight 2: Using a min-heap (priority queue) is crucial for optimally assigning ladders to the largest height differences. By allocating ladders to the largest climbs, we minimize brick usage for smaller climbs.

Which companies ask LeetCode 1642. Furthest Building You Can Reach in interviews?

This problem is frequently asked at DE Shaw, Dream11, Flipkart, Media.net, oyo.

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Furthest Building You Can Reach - LeetCode 1642 Solution

Furthest Building You Can Reach - Complete Solution Guide

Furthest Building You Can Reach is LeetCode problem 1642, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 l

Detailed Explanation

The problem presents a scenario where you're traversing a series of buildings with varying heights. You're given a limited number of bricks and ladders. When moving from one building to the next, if the next building is taller, you must either use bricks equal to the height difference or use a ladder. The goal is to determine the furthest building you can reach, starting from building 0, using an optimal strategy for utilizing bricks and ladders.

Solution Approach

The solution employs a greedy approach combined with a min-heap (priority queue). The heap stores the height differences that we've encountered so far. We iterate through the buildings, calculating the height difference between consecutive buildings. If the difference is positive, we push it onto the heap. If the number of elements in the heap exceeds the number of ladders we have, it implies we must use bricks for the smallest height difference in the heap. Therefore, we pop the smallest difference from the heap and deduct it from the number of bricks we have. If at any point our bricks become negative, it means we cannot proceed further, and we return the current building index. If we manage to traverse all the buildings, we return the last building index.

Step-by-Step Algorithm

Step 1: Initialize a min-heap `ladder_climbs` to store height differences.
Step 2: Iterate through the `heights` array from index 0 to `len(heights) - 2`.
Step 3: Calculate the height difference `diff` between `heights[i+1]` and `heights[i]`.
Step 4: If `diff` is positive, push `diff` onto the `ladder_climbs` heap.
Step 5: If the number of elements in `ladder_climbs` is greater than the number of `ladders`, pop the smallest element from the heap and subtract it from `bricks`.
Step 6: If `bricks` becomes negative, return the current building index `i`.
Step 7: If the loop completes without running out of bricks, return `len(heights) - 1`.

Key Insights

Insight 1: The problem is fundamentally about resource allocation and optimization. We need to decide when to use bricks and when to use ladders to maximize our progress.
Insight 2: Using a min-heap (priority queue) is crucial for optimally assigning ladders to the largest height differences. By allocating ladders to the largest climbs, we minimize brick usage for smaller climbs.
Insight 3: The core idea is to greedily allocate ladders to the largest height differences encountered so far. If we run out of ladders, we must use bricks. If we run out of both, we cannot proceed further.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Greedy, Heap (Priority Queue).

Companies

Asked at: DE Shaw, Dream11, Flipkart, Media.net, oyo.

Furthest Building You Can Reach | LeetCode 1642

Medium

Array Greedy Heap (Priority Queue)

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

Solution Code

Available in:

import java.util.PriorityQueue;

class Solution {
    public int furthestBuilding(int[] heights, int bricks, int ladders) {
        PriorityQueue<Integer> ladder_climbs = new PriorityQueue<>();

        for (int i = 0; i < heights.length - 1; i++) {
            int diff = heights[i + 1] - heights[i];

            if (diff > 0) {
                ladder_climbs.offer(diff);
                if (ladder_climbs.size() > ladders) {
                    bricks -= ladder_climbs.poll();
                }
                if (bricks < 0) {
                    return i;
                }
            }
        }

        return heights.length - 1;
    }
}

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Problem Info

Description

Problem 1642: Furthest Building You Can Reach - Detailed explanation and solution approaches.

Topics

Array Greedy Heap (Priority Queue)

Companies

DE ShawDream11FlipkartMedia.netoyo

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:1642

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize a min-heap `ladder_climbs` to store height differences.
2
Step 2: Iterate through the `heights` array from index 0 to `len(heights) - 2`.
3
Step 3: Calculate the height difference `diff` between `heights[i+1]` and `heights[i]`.
4
Step 4: If `diff` is positive, push `diff` onto the `ladder_climbs` heap.
5
Step 5: If the number of elements in `ladder_climbs` is greater than the number of `ladders`, pop the smallest element from the heap and subtract it from `bricks`.
6
Step 6: If `bricks` becomes negative, return the current building index `i`.
7
Step 7: If the loop completes without running out of bricks, return `len(heights) - 1`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant operation is the heap operations inside the loop. The loop runs 'n' times. The heap at max holds 'k' (number of ladders) elements. So each push or pop takes log(k) time giving O(n log k) as time complexity. Heap takes O(k) space.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Binary Search

Binary search to find the furthest building. For each potential furthest building, check if it's reachable within the given bricks and ladders constraints. This approach would involve simulating the building traversal process for each binary search iteration.

Time: O(n log n)Space: O(n) in worst case, potentially O(1) if done in-place

Trade-offs: While conceptually different, this approach has the same time complexity (n log n). It is generally less efficient in practice due to the overhead of simulating the traversal repeatedly in each binary search step.

✨

Key Insights

1Insight 1: The problem is fundamentally about resource allocation and optimization. We need to decide when to use bricks and when to use ladders to maximize our progress.
2Insight 2: Using a min-heap (priority queue) is crucial for optimally assigning ladders to the largest height differences. By allocating ladders to the largest climbs, we minimize brick usage for smaller climbs.
3Insight 3: The core idea is to greedily allocate ladders to the largest height differences encountered so far. If we run out of ladders, we must use bricks. If we run out of both, we cannot proceed further.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: `ladders` is 0. The algorithm still works correctly; the heap will never have more elements than ladders, so the condition `ladder_climbs.size() > ladders` will immediately trigger the brick usage.
Edge case 2: `bricks` is 0 and there are climbs needed. The algorithm correctly identifies that we can't proceed if we don't have enough bricks and returns the index accordingly.
Edge case 3: `heights` array is of length 1. The loop doesn't execute, and the algorithm returns index 0, which is correct.
Edge case 4: The input `heights` array is already in descending order (or equal heights). The `diff` will always be <= 0, so no ladders or bricks are used, and we can reach the end.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using a min-heap to prioritize ladder usage for the largest climbs. If you simply allocate ladders sequentially, you might end up using bricks for a large climb later when you could have saved them earlier.

✗Common mistake 2: Forgetting to handle the case when the number of ladders is zero. The code should still function correctly in this scenario, only using bricks.

✗Common mistake 3: Not considering the case when the height difference is non-positive (i.e., the next building is not taller). In this case, no bricks or ladders are needed, and you can proceed to the next building.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation in project management. Bricks and ladders are analogous to different types of resources with varying costs, and buildings are tasks. The goal is to complete the most tasks with limited resources.

🏦Real-world scenario 2: Network routing. Each building can represent a network node, and the height difference represents the cost (e.g., latency) of transmitting data between nodes. Bricks and ladders can represent different routing protocols. The goal is to reach the furthest node with the lowest possible cost.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 23, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Furthest Building You Can Reach"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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