🚨 Amazon OA Leaked! 2 Coding Problems - Not That Hard, But One Misunderstanding Ruins Everything

Amazon has recently launched another large-scale Online Assessment wave, with all OAs using the HackerRank platform. Based on oavoservice's coaching experience and candidate feedback, this batch of OA questions has a very stable structure, but places higher demands on candidates' modeling ability, data structure selection, and edge case handling.

Many candidates share a common sentiment after completing the OA:

The problems aren't particularly difficult, but once you misunderstand something, your code becomes almost impossible to salvage.

This article provides a systematic review of the two Coding problems from this Amazon OA round based on real feedback.

📊 Amazon OA Overview

Item	Details
Platform	HackerRank
Questions	2 Coding Problems
Difficulty	Medium to Hard
Key Focus	Complex business rule abstraction, dynamic data structures, greedy strategies, edge cases

✅ Problem 1: Server Instance Allocation & Cost Calculation

Problem Background

The system contains several servers, each with a certain number of idle instances. Server status is given as an array where the index represents the server ID and the value represents the current available instance count.

Now m customers arrive sequentially, and each customer needs to select a server to rent one instance. The server status changes continuously as customers make their selections.

Allocation & Cost Rules

For each customer, the system performs the following:

Among all current servers, select the one with the most idle instances
After successful selection, that server's idle instance count decreases by 1
Each selection produces a cost, calculated as:

cost = minimum idle instances (before selection) + maximum idle instances (before selection)

Final requirement: After m customers complete allocation, output the cumulative sum of all costs.

Core Problem Analysis

This is a very typical Amazon-style problem. It's not about complex algorithms, but about whether candidates can:

Quickly identify this as a dynamic min/max maintenance problem
Avoid inefficient full array scans in multiple update scenarios

🤯 Main Reasons for Point Loss

Not realizing cost must be calculated "before" selection
Scanning the entire array each round to find min/max, causing TLE
Not properly handling servers with instance count reduced to 0

oavoservice Perfect Solution

Use a heap structure to maintain server status:

import heapq

def calculateTotalCost(servers, m):
    if not servers or m == 0:
        return 0
    
    # Max heap (use negative values to simulate)
    max_heap = [-x for x in servers if x > 0]
    heapq.heapify(max_heap)
    
    # Maintain minimum value (can use Counter or sorted structure)
    from collections import Counter
    count = Counter(servers)
    
    total_cost = 0
    
    for _ in range(m):
        if not max_heap:
            break
        
        # Current max (before selection)
        current_max = -max_heap[0]
        # Current min (before selection)
        current_min = min(k for k, v in count.items() if v > 0)
        
        # Calculate cost (before selection)
        total_cost += current_max + current_min
        
        # Execute allocation: max server decreases by 1
        heapq.heappop(max_heap)
        count[current_max] -= 1
        
        new_val = current_max - 1
        if new_val > 0:
            heapq.heappush(max_heap, -new_val)
            count[new_val] += 1
    
    return total_cost

Complexity Analysis

Time: O(m log n), each heap operation is O(log n)
Space: O(n), storing heap and counter

✅ Problem 2: Parts Warehouse Distribution Maximization

Problem Background

The system contains multiple log deliveries, each log corresponding to a certain number of parts. There are k warehouses, where k is an even number.

Storage Rules

Each warehouse can only store parts from the same log
Parts from the same log can be distributed across multiple warehouses
Some log parts may not be stored in any warehouse

Sorting & Target Constraints

After all warehouses complete storage:

Sort warehouses by their stored parts count
The top k/2 warehouses are considered "the half that stores the most"
The bottom k/2 warehouses are considered "the half that stores the least"

Requirement: Output the maximum possible sum of parts in the bottom half (k/2 warehouses that store the least).

🤯 Core Difficulty

The difficulty isn't in implementation, but in understanding the objective.

Many candidates instinctively want to "store as many parts as possible," but the real optimization target isn't total storage, but rather:

After sorting, maximize the sum of the bottom half warehouses.

This means:

Don't let the top half warehouses consume too many parts
Distribution strategy must be as balanced as possible, while satisfying the single-log-per-warehouse constraint

Common Mistakes

Ignoring the single log source per warehouse constraint
Only focusing on total parts count, not considering post-sort distribution
Assuming all parts must be allocated, causing bottom warehouses to be suppressed

oavoservice Solution Approach

def maxBottomHalfSum(logs, k):
    """
    logs: array of parts count for each log
    k: number of warehouses (even)
    Goal: maximize the sum of bottom k/2 warehouses after sorting
    """
    # Key insight: to maximize bottom half, distribution should be balanced
    # Use binary search to find optimal "minimum allocation value"
    
    def canAchieve(min_val):
        """Check if all warehouses can store at least min_val parts"""
        warehouses_filled = 0
        for log in logs:
            # Number of warehouses each log can fill
            warehouses_filled += log // min_val
        return warehouses_filled >= k
    
    # Binary search for maximum min_val
    left, right = 1, max(logs) if logs else 0
    result = 0
    
    while left <= right:
        mid = (left + right) // 2
        if canAchieve(mid):
            result = mid
            left = mid + 1
        else:
            right = mid - 1
    
    # Sum of bottom k/2 warehouses
    return result * (k // 2)

Note: This is a simplified approach. Actual problems may have additional constraints requiring adjustment.

💡 Amazon OA Real Style Summary

From these two problems, we can see Amazon OA's focus areas:

Dimension	Specifics
Business Modeling	Long problem descriptions, need to quickly abstract core logic
Data Structures	Flexible use of heaps, priority queues, hash tables
Edge Cases	Empty arrays, zero values, extreme cases
Optimization Target	Clearly understand "what exactly to maximize/minimize"

🎯 Why Do Many Choose oavoservice?

In oavoservice's real cases, many candidates' problems aren't "completely unable to code," but rather:

Insufficient OA time, no room for trial and error
Complex rules, easy to trip on details
Strict platform restrictions, one mistake means elimination

Amazon OA's nature is screening, not teaching. In such situations, choosing stable, compliant, and experienced professional support is often more effective than simply "practicing a few more problems."

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