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Neon Number

Java coding interview problem for Number Logic: Neon Number.

Checking whether a number is a Neon Number is one of the easiest and most commonly asked beginner Java coding interview questions.

This problem helps interviewers evaluate your understanding of:

  • Arithmetic operations
  • Loops
  • Digit extraction
  • Mathematical logic
  • Problem-solving skills

Neon Number problems are frequently used in beginner programming interviews and coding practice platforms.

What is a Neon Number?

A Neon Number is a number whose sum of the digits of its square is equal to the original number.

In simple words,

Neon Number

=

Sum of Digits of (Number²)

Mathematical Definition

If

N

is a number,

then

Square = N × N

If

Sum of Digits(Square)

=

N

then

N is a Neon Number

Example 1

Number

9

Square

9 × 9

=

81

Digits

8

1

Sum

8 + 1

=

9

Since

9 = 9

Therefore

9 is a Neon Number

Example 2

Number

1

Square

1

Digits

1

Sum

1

Therefore

1 is a Neon Number

Example 3

Number

12

Square

144

Digits

1

4

4

Sum

1 + 4 + 4

=

9

Since

9 ≠ 12

Therefore

12 is NOT a Neon Number

Some Neon Numbers

Number Square Digit Sum Neon Number?
0 0 0 ✅ Yes
1 1 1 ✅ Yes
9 81 9 ✅ Yes
10 100 1 ❌ No
12 144 9 ❌ No
15 225 9 ❌ No

Real Interview Question

Write a Java program to check whether a given number is a Neon Number.

Understanding the Logic

Suppose

Number = 9

First,

calculate its square.

9² = 81

Extract every digit.

8

1

Find their sum.

8 + 1 = 9

Compare the sum with the original number.

If both are equal,

the number is a Neon Number.

Visual Representation

Input

9

Processing

9

↓

Square

↓

81

↓

Extract Digits

↓

8

↓

1

↓

Sum

↓

8 + 1

↓

9

↓

Compare

↓

9 == 9

↓

Neon Number

Brute Force Approach

The simplest solution is

  • Store the original number.
  • Find its square.
  • Extract every digit of the square.
  • Add the digits.
  • Compare the sum with the original number.

Algorithm

Step 1

Read the number.

Step 2

Store the original number.

original = number;

Step 3

Find the square.

square = number * number;

Step 4

Initialize

sum = 0;

Step 5

Extract the last digit.

digit = square % 10;

Step 6

Add the digit to the sum.

Step 7

Remove the last digit.

square = square / 10;

Step 8

Repeat until

square = 0

Step 9

Compare

sum == original

Dry Run

Input

9

Square

81

Initial

sum = 0
Digit Running Sum
1 1
8 9

Comparison

9 == 9

Output

Neon Number

Another Dry Run

Input

12

Square

144

Initial

sum = 0
Digit Running Sum
4 4
4 8
1 9

Comparison

9 ≠ 12

Output

Not a Neon Number

Approach 1 — Using Iteration (Loop)

This is the easiest and most common interview solution.

Complete Java Program

public class NeonNumber {

    public static void main(String[] args) {

        int number = 9;

        int original = number;

        int square = number * number;

        int sum = 0;

        while (square > 0) {

            int digit = square % 10;

            sum += digit;

            square /= 10;

        }

        if (sum == original) {

            System.out.println(original + " is a Neon Number");

        } else {

            System.out.println(original + " is NOT a Neon Number");

        }

    }

}

Output

9 is a Neon Number

Step-by-Step Code Explanation

Step 1

Declare the number.

int number = 9;

Current value

9

Step 2

Store the original number.

int original = number;

This is required because the value of

square

changes while extracting digits.

Step 3

Find the square.

int square = number * number;

Result

81

Step 4

Initialize the sum.

int sum = 0;

Initially

sum = 0

Step 5

Extract the last digit.

int digit = square % 10;

First digit

1

Step 6

Add the digit.

sum += digit;

Running sum

1

↓

9

Step 7

Remove the last digit.

square /= 10;

Values become

81

↓

8

↓

0

Step 8

Compare

sum == original

If true,

the number is a Neon Number.

Otherwise,

it is not.

Example Execution

Input

Number = 1

Square

1

Digit Sum

1

Output

1 is a Neon Number

Input

Number = 15

Square

225

Digit Sum

2 + 2 + 5

=

9

Output

15 is NOT a Neon Number

Why Does This Work?

The algorithm first computes the square of the number.

Then it extracts each digit of the square using the modulus (%) operator, adds the digits together, and finally compares the sum with the original number.

If both values are equal, the number satisfies the mathematical definition of a Neon Number.

Advantages of This Approach

  • Very easy to understand.
  • Uses simple arithmetic operations.
  • Demonstrates digit extraction.
  • Excellent for beginners.
  • Frequently asked in entry-level interviews.

Drawbacks

Although this solution is straightforward, interviewers often ask follow-up questions such as:

  • Can you create a reusable method?
  • How would you check Neon Numbers in a range?
  • What is the time complexity?
  • Can you optimize the digit-sum calculation?
  • How does a Neon Number differ from an Armstrong Number or Strong Number?

In Part 2, we'll cover:

  • Reusable Method Approach
  • Printing Neon Numbers in a Range
  • Time and Space Complexity
  • Comparison of Approaches
  • Common Interview Mistakes
  • Interview Follow-up Questions
  • Related Coding Problems
  • Key Takeaways
  • Interview Tips

Approach 2 — Using a Reusable Method

Instead of writing the Neon Number logic inside the main() method, we can create a reusable method.

This approach improves:

  • Code readability
  • Maintainability
  • Reusability
  • Unit testing

Java Program

public class NeonNumberMethod {

    static boolean isNeon(int number) {

        int square = number * number;

        int sum = 0;

        while (square > 0) {

            int digit = square % 10;

            sum += digit;

            square /= 10;

        }

        return sum == number;

    }

    public static void main(String[] args) {

        int number = 9;

        if (isNeon(number)) {

            System.out.println(number + " is a Neon Number");

        } else {

            System.out.println(number + " is NOT a Neon Number");

        }

    }

}

Output

9 is a Neon Number

Approach 3 — Print Neon Numbers in a Range

Sometimes interviewers ask:

Print all Neon Numbers between 1 and N.

The solution is to reuse the isNeon() method.

Java Program

public class NeonNumbersRange {

    static boolean isNeon(int number) {

        int square = number * number;

        int sum = 0;

        while (square > 0) {

            sum += square % 10;

            square /= 10;

        }

        return sum == number;

    }

    public static void main(String[] args) {

        int limit = 100;

        System.out.println("Neon Numbers:");

        for (int i = 0; i <= limit; i++) {

            if (isNeon(i)) {

                System.out.print(i + " ");

            }

        }

    }

}

Output

Neon Numbers:

0 1 9

Dry Run (Reusable Method)

Input

Number = 9

Processing

Square

↓

81

↓

Digit

1

↓

Sum = 1

↓

Digit

8

↓

Sum = 9

↓

Compare

9 == 9

↓

Return true

Time Complexity

Suppose the square contains

d

digits.

Basic Iterative Solution

Operation Complexity
Time O(d)
Space O(1)

Reusable Method

Operation Complexity
Time O(d)
Space O(1)

Printing Neon Numbers in a Range

Operation Complexity
Time O(n × d)
Space O(1)

where

n

is the upper limit.

Comparison of Approaches

Approach Time Space Recommended
Basic Loop O(d) O(1) Good for Beginners
Reusable Method O(d) O(1) ✅ Best for Interviews
Range Solution O(n × d) O(1) Useful Follow-up

Common Mistakes

Mistake 1

Comparing with the square instead of the original number.

Wrong

sum == square

Correct

sum == number

Mistake 2

Forgetting to store the original number.

Wrong

number

may change during processing.

Correct

int original = number;

Mistake 3

Using multiplication instead of addition.

Wrong

sum *= digit;

Correct

sum += digit;

Mistake 4

Not removing the processed digit.

Wrong

while(square > 0) {

    sum += square % 10;

}

Correct

square /= 10;

Otherwise,

the loop never ends.

Mistake 5

Ignoring

0

The square of

0

is

0

Digit sum

0

Therefore,

0 is also a Neon Number.

Interview Follow-up Questions

Q1. What is a Neon Number?

Q2. Why is 9 a Neon Number?

Q3. Is 1 a Neon Number?

Q4. Is 0 a Neon Number?

Q5. Print all Neon Numbers in a range.

Q6. What is the time complexity?

Q7. Can negative numbers be Neon Numbers?

Q8. Explain digit extraction.

Q9. Compare Neon Number and Armstrong Number.

Q10. Write a reusable method for checking Neon Numbers.

Related Coding Problems

  • Armstrong Number
  • Strong Number
  • Perfect Number
  • Palindrome Number
  • Reverse Number
  • Sum of Digits
  • Happy Number
  • Automorphic Number

Key Takeaways

  • A Neon Number is a number whose sum of the digits of its square equals the original number.
  • The first Neon Numbers are:
0

1

9
  • Calculate the square first.
  • Extract digits using:
digit = square % 10;
  • Remove digits using:
square /= 10;
  • Add every digit to a running sum.
  • Compare the final sum with the original number.
  • The algorithm runs in O(d) time and O(1) extra space, where d is the number of digits in the square.

Frequently Asked Interview Questions

Q1. What is a Neon Number?

A Neon Number is a number whose sum of the digits of its square equals the original number.

Example

9² = 81

8 + 1 = 9

Q2. Is every single-digit number a Neon Number?

No.

Only

0

1

9

are Neon Numbers.

Q3. Can a negative number be a Neon Number?

Generally, No.

Neon Numbers are defined only for non-negative integers.

Q4. What is the difference between a Neon Number and a Strong Number?

Neon Number Strong Number
Uses the square of the number Uses the factorial of each digit
Adds digits of the square Adds factorials of digits
Example: 9 Example: 145

Q5. Why is this problem asked in interviews?

Because it tests:

  • Arithmetic operations
  • Digit extraction
  • Looping concepts
  • Problem decomposition
  • Basic mathematical reasoning

Interview Tip

If an interviewer asks:

"Write a Java program to check whether a number is a Neon Number."

Start by explaining the mathematical definition:

A number is a Neon Number if the sum of the digits of its square equals the original number.

Implement the straightforward loop-based solution by computing the square, extracting its digits using % and /, and summing them. Mention that the algorithm runs in O(d) time, where d is the number of digits in the square, and uses O(1) extra space. As a follow-up, demonstrate good coding practices by moving the logic into a reusable isNeon() method and discuss how it can be extended to print all Neon Numbers within a given range.