🔥 NVIDIA 2026 OA Real Questions Leaked! Seemingly Simple Array Problem - Meeting Senior Interviewers is the Real "Nightmare"!

Breaking the Illusion: Manufacturing Anxiety

Recently, NVIDIA's OA tests are being distributed extensively.

Many candidates' first reaction upon seeing the question: "Easy! This is just basic array processing, right?"

Wrong.

If you treat this as a simple HashMap + sorting problem, even if you pass all Test Cases, in the interviewer's eyes it's just toy code.

Our oavoservice team obtained the real questions immediately and discovered countless hidden traps involving data consistency and edge case handling.

Question Breakdown: Demonstrating Expertise

🔍 Core Question: Restore Original Array from Doubled Shuffled Array

Problem Description: An integer array original is transformed into a changed array by:

1. Appending twice the value of every element in original to the array
2. Randomly shuffling the resulting array

Given array changed, return original if it's a valid doubled array; otherwise return an empty array.

Key Challenges:

• Frequency counting boundary handling
• Zero value special cases
• Time complexity optimization

Input/Output Example:

Input: changed = [1,3,4,2,6,8]
Output: [1,3,4] 
Explanation: original=[1,3,4], changed=[1,3,4,2,6,8]

oavoservice Exclusive Analysis: This problem disguises itself as a basic array problem while actually testing data structure design and algorithm complexity optimization.

Fatal Pitfalls: Why Others Fail

🤯 Hidden Killer: Zero Value Trap + Frequency Pairing Logic

This is where most candidates fail.

Many students are accustomed to hard coding thinking and directly use HashMap:

# Wrong rote memorization approach
def findOriginalArray(changed):
    if len(changed) % 2 != 0:
        return []
    
    count = {}
    for num in changed:
        count[num] = count.get(num, 0) + 1
    
    result = []
    for num in sorted(changed):
        if count[num] > 0 and count.get(num * 2, 0) > 0:
            result.append(num)
            count[num] -= 1
            count[num * 2] -= 1
    
    return result if len(result) == len(changed) // 2 else []

But in NVIDIA interviewers' eyes, this means:

1. Corner Case blind spots: When x = 0, you need an even number of zeros to pair
2. Shallow algorithmic understanding: Didn't consider the necessity of small-to-large pairing
3. Lack of engineering judgment: Poor code readability, high maintenance costs

The interviewer's follow-up questions are often not because you couldn't solve it, but because you lack trade-off thinking.

Correct Industrial-Grade Solution

def findOriginalArray(changed):
    n = len(changed)
    if n % 2 == 1:
        return []
    
    from collections import Counter
    count = Counter(changed)
    
    # Special handling for zero values
    if count[0] % 2 == 1:
        return []
    
    result = []
    # Pair from small to large to avoid double consumption
    for x in sorted(count.keys()):
        if count[x] == 0:
            continue
            
        if x == 0:
            # Zero values need to appear in pairs
            result.extend([0] * (count[x] // 2))
        else:
            # Check if there are enough 2x to pair with
            if count[x] > count[2 * x]:
                return []
            result.extend([x] * count[x])
            count[2 * x] -= count[x]
        
        count[x] = 0
    
    return result

Conversion Closing: Service Upgrade

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