Java Program to Check If a Number is Neon Number or Not
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 :
- 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 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.
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)
