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C Program To Check Neon Number

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  • Difficulty Level : Hard
  • Last Updated : 26 Sep, 2022
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Given a number (num)  we need to check whether it is a Neon Number ( i.e. a number where the sum of digits of the square of the number is equal to the number ) and return “true” or “false” accordingly.

Example:

Input: num = 9

Output: true

Explanation: square of 9 is 9 * 9 = 81 , sum of digit of square is 8 + 1 = 9 (i.e equal to given number).

Input: num = 10
Output: false

Explanation: Square of 10 is 10 * 10 = 100 , sum of digit of square is 1 + 0 + 0 = 1 (i.e. not equal to given number).

Approach

The basic idea is to first calculate the square of the number and then by calculating the sum of digits of the square we can check whether the given number is Neon Number or not.

Algorithm

  • At first, we need to find the square of the given number.
  • Then we need to extract the digits of the square of the number using a loop.
  • Then we need to calculate the sum of digits.
  • At last, we need to check the sum with the given number, and the Outcome will be:
    • ” true ” , if the sum is equal to the given number, or
    • ” false ” , if the sum is not equal to the given number.

Example:

C++




// C program to demonstrate whether
// a number is Neon number or not
#include <stdio.h>
 
int Check_Neon_Number(int num) {
 
    // Calculating the square of the number
    int square = num * num;
 
    // Copying the square in a variable
    // to extract the digit
    int n = square;
 
    // Declaring a variable to store the digits
    int digit;
 
    // Initializing a variable to
    // calculate the sum of digits
    int sum = 0;
 
    // To calculate the sum of digits
    while (n != 0) {
 
        // Extracting the digit
        digit = n % 10;
        sum = sum + digit;
        n = n / 10;
    }
 
    // Checking the condition of a Neon Number
    if (sum == num)
        return 1; // If condition is true.
    else
        return 0; // If condition is false.
}
 
// Driver Code
int main()
{
    int num = 9;
 
    // Calling the function
    int ans = Check_Neon_Number(num);
    if (ans == 1)
 
        // The number is Neon
        printf("true");
    else
 
        // The number is not Neon
        printf("false");
    return 0;
}

Output

true

Time Complexity: O(log n)

Auxiliary Space: O(1) as no extra space is required, so space is constant


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