A neon number is a number where the sum of digits of the square of the number is equal to the number. The task is to check and print neon numbers in a range.
Illustration:
Case 1:
Input : 9
Output : Given number 9 is Neon number
Explanation : square of 9=9*9=81;
sum of digit of square : 8+1=9(which is equal to given number)
Case 2:
Input : 8
Output : Given number is not a Neon number
Explanation : square of 8=8*8=64
sum of digit of square : 6+4=10(which is not equal to given number)
Algorithm :
- First, find the square of the given number.
- Find the sum of the digit of the square by using a loop.
 - The condition checksum is equal to the given number
- Return true
- Else return false.
Pseudo code : Square =n*n;
while(square>0)
{
int r=square%10;
sum+=r;
square=square/10;
}
Example:
Java
// Java Program to Check If a Number is Neon number or notÂ
// Importing java input/output libraryimport java.io.*;Â
class GFG {Â
    // Method to check whether number is neon or not    // Boolean type    public static boolean checkNeon(int n)    {        // squaring the number to be checked        int square = n * n;Â
        // Initializing current sum to 0        int sum = 0;Â
        // If product is positive        while (square > 0) {Â
            // Step 1: Find remainder            int r = square % 10;Â
            // Add remainder to the current sum            sum += r;Â
            // Drop last digit of the product            // and store the number            square = square / 10;        }Â
        // Condition check        // Sum of digits of number obtained is        // equal to original number        if (sum == n)Â
            // number is neon            return true;        elseÂ
            // number is not neon            return false;    }Â
    // Main driver method    public static void main(String[] args)    {        // Custom input        int n = 9;Â
        // Calling above function to check custom number or        // if user entered number via Scanner class        if (checkNeon(n))Â
            // Print number considered is neon            System.out.println("Given number " + n                               + " is Neon number");        elseÂ
            // Print number considered is not neon            System.out.println("Given number " + n                               + " is not a Neon number");    }} |
Output
Given number 9 is Neon number
Time Complexity: O(l) where l is the number of the digit in the square of the given number
Recursive Approach:Â
Explanation:
- In this approach, we use a recursive function isNeonNumber to check if the input number is a neon number.
- The function takes two arguments: the square of the input number and the input number itself.
- At each recursive call, we extract the last digit of the square number and subtract it from the input number.
- We then call the function recursively with the remaining digits of the square number and the updated input number.
- If the square number has no more digits left (i.e., square == 0), then we check if the input number is zero (i.e., number == 0).
If the input number is zero, then the original input number is a neon number; otherwise, it is not.
Java
import java.util.Scanner;Â
public class NeonNumber {Â Â Â Â public static void main(String[] args) {Â Â Â Â Â Â Â Â int number=9;Â Â Â Â Â Â Â Â int square = number * number;Â Â Â Â Â Â Â Â if (isNeonNumber(square, number)) {Â Â Â Â Â Â Â Â Â Â Â Â System.out.println(number + " is a neon number");Â Â Â Â Â Â Â Â } else {Â Â Â Â Â Â Â Â Â Â Â Â System.out.println(number + " is not a neon number");Â Â Â Â Â Â Â Â }Â Â Â Â }Â
    private static boolean isNeonNumber(int square, int number) {        if (square == 0) {            return number == 0;        } else {            int digit = square % 10;            return isNeonNumber(square / 10, number - digit);        }    }} |
Output
9 is a neon number
 Time Complexity: O(logn)
 Auxiliary Space: O(logn)
