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 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)