My Airbnb HackerRank Experience: Disk Clustering Question and Prep Notes

This is a rewritten walkthrough of an Airbnb HackerRank online assessment experience focused on a single coding problem about disk sector clustering. The original assessment described here had one question, a 60-minute limit, and a difficulty that felt medium-high because the key insight was mathematical rather than implementation-heavy. The goal of this article is to preserve the useful interview-prep value: how to understand the problem, derive the greedy solution, manage time, and review edge cases responsibly.

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Airbnb HackerRank Question Overview

The assessment question was based on a disk clustering scenario. You are given a list of sector costs and must split the sectors into a fixed number of contiguous clusters. Each cluster has a cost based on the first and last sector in that cluster. The task is to compute the minimum and maximum possible total clustering cost.

Problem Inputs

• n is the total number of disk sectors.
• sectorCost[i] is the cost of the sector at index i.
• clusterCount is the number of clusters that must be formed.

Cluster Cost Rule

For a cluster spanning sectors from x through y, the cluster cost is:

clusterCost = sectorCost[x] + sectorCost[y]

The clusters must cover the array in order, and each cluster is a contiguous segment. That means the main decision is where to place the clusterCount - 1 split points.

Core Insight Behind the Solution

The first and last sectors of the entire array always contribute to the total cost. Those two values are fixed no matter where the internal splits are placed.

Every internal split between positions i and i + 1 adds two boundary values to the total cost:

sectorCost[i] + sectorCost[i + 1]

Therefore, the total cost can be modeled as:

totalCost = sectorCost[0] + sectorCost[n - 1] + sum(selected adjacent split costs)

The selected split costs are the costs of the clusterCount - 1 internal boundaries. Once this formula is clear, the problem becomes much simpler. You do not need dynamic programming because choosing one split boundary does not change the value of another split boundary.

Solution Approach

Step 1: Build the Adjacent Pair Cost Array

Create a new array containing the cost of placing a split between every adjacent pair of sectors.

pairCost[i] = sectorCost[i] + sectorCost[i + 1]

There are n - 1 possible split positions. You only need to choose clusterCount - 1 of them.

Step 2: Compute the Minimum Cost

To minimize the total cost, choose the smallest clusterCount - 1 values from pairCost.

1. Sort pairCost in ascending order.
2. Sum the first clusterCount - 1 values.
3. Add sectorCost[0] and sectorCost[n - 1].

The formula is:

minimumCost = sectorCost[0] + sectorCost[n - 1] + sum(smallest split costs)

Step 3: Compute the Maximum Cost

To maximize the total cost, choose the largest clusterCount - 1 values from pairCost.

1. Use the sorted pairCost array.
2. Sum the last clusterCount - 1 values.
3. Add sectorCost[0] and sectorCost[n - 1].

The formula is:

maximumCost = sectorCost[0] + sectorCost[n - 1] + sum(largest split costs)

Step 4: Return Both Values

The final result is usually returned as a two-element array:

[minimumCost, maximumCost]

The overall time complexity is O(n log n) because of sorting. The auxiliary space complexity is O(n) for the adjacent pair cost array.

Example Implementation Pattern

Here is the language-agnostic structure I would use in an assessment.

build pairCost from all adjacent sector pairs
sort pairCost
baseCost = first sector cost + last sector cost
splitsNeeded = clusterCount - 1
minimumCost = baseCost + sum of the first splitsNeeded pair costs
maximumCost = baseCost + sum of the last splitsNeeded pair costs
return [minimumCost, maximumCost]

For Python, the implementation can stay compact.

def disk_clustering_cost(sector_cost, cluster_count):
    n = len(sector_cost)
    pair_cost = [sector_cost[i] + sector_cost[i + 1] for i in range(n - 1)]
    pair_cost.sort()

    base_cost = sector_cost[0] + sector_cost[-1]
    splits_needed = cluster_count - 1

    min_cost = base_cost + sum(pair_cost[:splits_needed])
    max_cost = base_cost + sum(pair_cost[-splits_needed:]) if splits_needed else base_cost

    return [min_cost, max_cost]

A small implementation detail matters here. When cluster_count is 1, there are no internal splits, so both minimum and maximum cost are just the first plus last sector cost.

Why the Greedy Strategy Works

The split boundaries are independent. A split between sectors i and i + 1 always contributes exactly sectorCost[i] + sectorCost[i + 1]. That contribution does not depend on which other split points are chosen.

Because all candidate split costs can be evaluated independently, minimizing the total means selecting the smallest boundary costs. Maximizing the total means selecting the largest boundary costs. This is why sorting is enough and why a more complex interval DP solution would be unnecessary.

Time Management for a 60-Minute Assessment

For this type of Airbnb HackerRank question, I would divide the hour into four phases.

Phase 1: Read and Model the Problem

Spend about 15 minutes reading the prompt, checking the input format, and writing down what a cluster contributes to the total cost. Manually compute at least one sample so you can see which values are fixed and which values change.

Phase 2: Derive and Validate the Formula

Spend about 20 minutes testing possible split placements on small arrays. The key is to notice that the first and last array values are fixed, while each internal split contributes an adjacent-pair sum. Once you see that pattern, validate it against a few different cluster counts.

Phase 3: Implement Clearly

Spend about 15 minutes coding the solution with clean variable names. Keep the implementation simple enough that you can reason about boundary cases without rereading complex logic.

Phase 4: Test and Polish

Reserve the final 10 minutes for edge cases, syntax checks, and performance review. Do not rely only on visible sample tests. Hidden tests often expose assumptions about minimum sizes, single clusters, and unusual values.

Edge Cases to Test

These are the cases I would test before submitting.

• clusterCount = 1, where the answer should be the cost of the whole array as one cluster.
• clusterCount = n, where every sector is its own cluster.
• Very small arrays that still satisfy the input constraints.
• Repeated sector costs, where many split choices may tie.
• Zero or negative sector costs, if the problem constraints allow them or do not clearly exclude them.
• Large arrays, to confirm that the O(n log n) solution avoids timeouts.

What the Question Evaluates

Problem Decomposition

The disk-sector story is mostly a wrapper around interval partitioning. Strong candidates identify the real structure beneath the story and separate fixed cost from variable cost.

Mathematical Modeling

The hardest part is not the code. The hardest part is deriving the total-cost formula and realizing that only adjacent split costs need to be optimized.

Greedy Reasoning

This problem rewards candidates who can prove that a greedy choice is valid. The proof comes from split independence: each split has a fixed contribution no matter where other splits are placed.

Code Reliability

Clean handling of clusterCount = 1, slicing behavior, and indexing prevents avoidable hidden-test failures. A short solution is valuable only if the boundary cases are correct.

Airbnb HackerRank Preparation Strategy

Practice Greedy Cost Problems

Prioritize greedy problems where the main task is to transform a story into a cost formula. Look for patterns involving adjacent pairs, boundary costs, sorting, and selecting the smallest or largest k values.

Practice Interval and Partitioning Problems

This problem looks like an interval partitioning problem, even though the final solution is simpler than many DP partition problems. Practice enough interval problems to recognize when dynamic programming is necessary and when the variables are independent.

Build a Personal Edge-Case Checklist

Before every timed assessment, have a checklist for single-element behavior, smallest valid input, largest valid input, ties, zeros, negatives, and off-by-one indexing. A checklist helps you stay calm when the prompt is unfamiliar.

Simulate the Real Timer

Run 60-minute simulations using one or two medium-level coding problems. Follow the same workflow every time: read, model, derive, implement, test, and review. This builds the habit of reserving time for hidden-test thinking instead of spending the full hour coding.

Review Mistakes by Pattern

After practice, do not only record whether you solved the problem. Record why you missed the insight, which constraints you overlooked, and which edge cases failed. That kind of review turns one problem into a reusable interview pattern.

Using ExtraBrain Responsibly for This Kind of Prep

ExtraBrain is a free, local-first Mac desktop AI interview assistant and meeting copilot with live transcription, screen-aware context, bring-your-own AI providers, local Gemma 4 where installed and compatible, and privacy controls. For preparation, it can help you review prompts, structure explanations, practice thinking aloud, and turn mistakes into follow-up study notes.

ExtraBrain can also work as a focused AI second brain for interviews and meetings by organizing live sessions, transcripts, notes, screen context, and review. If you use it during any assessment or interview, only do so where the rules explicitly allow AI assistance, transcription, screenshots, or notes. When configured with local Parakeet transcription and local Gemma 4 on compatible hardware, ExtraBrain can support a more local workflow, while external providers may receive selected prompts, transcript text, screenshots, audio, or context depending on your configuration.

FAQ

Does Airbnb’s HackerRank test require a webcam?

In the experience summarized here, the HackerRank assessment did not require a webcam. Assessment requirements can vary by role, region, year, and invitation, so always follow the exact instructions in your own test invite.

What was the main trick in the disk clustering question?

The main trick was realizing that the first and last sector costs are fixed. Every internal split only adds the cost of one adjacent pair. After that, the minimum and maximum answers come from sorting adjacent pair costs.

Is dynamic programming needed for this problem?

No, not for the version described here. Because each split contributes independently, sorting adjacent pair costs is enough.

How should I use spare time after coding?

First, test the smallest valid cases and the clusterCount = 1 case. Second, test the case where every sector becomes its own cluster. Third, review indexing and slicing carefully because most mistakes in this solution are off-by-one errors. Finally, recheck the complexity to confirm it can handle large inputs.

Can ExtraBrain generate interview answer outlines?

ExtraBrain can help generate answer outlines, STAR structures, technical explanations, and follow-up questions from live transcript and screen context. Candidates remain responsible for honest and allowed use.

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