C Program To Check Neon Number
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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