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
- Return true
- Else return false.
Pseudo code : Square =n*n;
while(square>0)
{
int r=square%10;
sum+=r;
square=square/10;
}
Example:
Java
import java.io.*;
class GFG {
public static boolean checkNeon( int n)
{
int square = n * n;
int sum = 0 ;
while (square > 0 ) {
int r = square % 10 ;
sum += r;
square = square / 10 ;
}
if (sum == n)
return true ;
else
return false ;
}
public static void main(String[] args)
{
int n = 9 ;
if (checkNeon(n))
System.out.println( "Given number " + n
+ " is Neon number" );
else
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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