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

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  • Difficulty Level : Easy
  • Last Updated : 16 Mar, 2023
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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 :

  1. First, find the square of the given number.
  2. Find the sum of the digit of the square by using a loop.
     
  3. 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);
        }
    }
}

Output

9 is a neon number

 Time Complexity: O(logn)

 Auxiliary Space: O(logn)


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