Java Program to Check If a Number is Neon Number or Not

Last Updated : 16 Mar, 2023

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
1. Return true
2. 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 not // Importing java input/output library import java.io.*; class GFG {  // 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;  // Initializing current sum to 0  int sum = 0;  // If product is positive  while (square > 0) {  // Step 1: Find remainder  int r = square % 10;  // Add remainder to the current sum  sum += r;  // Drop last digit of the product  // and store the number  square = square / 10;  }  // Condition check  // Sum of digits of number obtained is  // equal to original number  if (sum == n)  // number is neon  return true;  else  // number is not neon  return false;  }  // Main driver method  public static void main(String[] args)  {  // Custom input  int n = 9;  // Calling above function to check custom number or  // if user entered number via Scanner class  if (checkNeon(n))  // Print number considered is neon  System.out.println("Given number " + n  + " is Neon number");  else  // 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:

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; public class NeonNumber {  public static void main(String[] args) {  int number=9;  int square = number * number;  if (isNeonNumber(square, number)) {  System.out.println(number + " is a neon number");  } else {  System.out.println(number + " is not a neon number");  }  }  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);  }  } }