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 libraryimport 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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