A number is said to be a Peterson number if the sum of factorials of each digit of the number is equal to the number itself.
Example:
Input : n = 145 Output = Yes Explanation: 145 = 5! + 4! + 1! = 120 + 24 +1 = 145 Input : n = 55 Output : No Explanation: 5! + 5! = 120 + 120 = 240 Since 55 is not equal to 240 It is not a Peterson number.
We will pick each digit (Starting from the last digit) of the given number and find its factorial. And add all factorials. Finally, we check if the sum of factorials is equal to number or not.
C++
// C++ program to determine whether the number is // Peterson number or not #include <iostream> using namespace std; // To quickly find factorial of digits int fact[10] = { 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880 }; // Function to check if a number is Peterson // or not bool peterson( int n) { int num = n, sum = 0; // stores the sum of factorials of // each digit of the number. while (n > 0) { int digit = n % 10; sum += fact[digit]; n = n / 10; } // Condition check for a number to // be a Peterson Number return (sum == num); } // Driver Program int main() { int n = 145; if (peterson(n)) cout << "Yes" ; else cout << "No" ; return 0; } |
Java
//checks whether a number entered by user is peterson number or not import java.util.*; class GFG { public static void main(String args[]) { Scanner sc= new Scanner(System.in); //taking input from the user System.out.println( "Enter the number" ); int num=sc.nextInt(); int temp=num; //storing the number in a temporary variable int f= 1 ,sum= 0 ; while (num!= 0 ) //running while loop until number becomes zero { f= 1 ; //extracting last digit of the number //and storing in r int r=num% 10 ; //for loop to find the factorial of a digit for ( int i= 1 ;i<=r;i++) { f=f*i; } sum=sum+f; //adding the factotial of the digits num=num/ 10 ; } //checking if the sum of the factorial of digits //is equal to the number or not if (sum==temp) System.out.println( "PETERSON NUMBER" ); else System.out.println( "NOT PETERSON NUMBER" ); } } |
Python3
# Python3 code to determine whether the # number is Peterson number or not # To quickly find factorial of digits fact = [ 1 , 1 , 2 , 6 , 24 , 120 , 720 , 5040 , 40320 , 362880 ] # Function to check if a number # is Peterson or not def peterson(n): num = n sum = 0 # stores the sum of factorials of # each digit of the number. while n > 0 : digit = int (n % 10 ) sum + = fact[digit] n = int (n / 10 ) # Condition check for a number # to be a Peterson Number return ( sum = = num) # Driver Code n = 145 print ( "Yes" if peterson(n) else "No" ) # This code is contributed by "Sharad_Bhardwaj".. |
C#
// C# program to determine whether the // number is Peterson number or not using System; public class GFG { // To quickly find factorial of digits static int [] fact = new int [10] { 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880 }; // Function to check if a number is // Peterson or not static bool peterson( int n) { int num = n; int sum = 0; // stores the sum of factorials of // each digit of the number. while (n > 0) { int digit = n % 10; sum += fact[digit]; n = n / 10; } // Condition check for a number to // be a Peterson Number return (sum == num); } // Driver Program static public void Main() { int n = 145; if (peterson(n)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } } // This code is contributed by vt_m. |
PHP
<?php // PHP program to determine // whether the number is // Peterson number or not // To quickly find // factorial of digits $fact = array (1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880); // Function to check if // a number is Peterson // or not function peterson( $n ) { $num = $n ; $sum = 0; // stores the sum of factorials of // each digit of the number. while ( $n > 0) { $digit = $n % 10; $n = $n / 10; } // Condition check for // a number to be a // Peterson Number return ( $sum == $num ); } // Driver Code $n = 145; if (peterson( $n )) echo "Yes" ; else echo "No" ; // This code is contributed by ajit ?> |
Javascript
<script> // Javascript program to determine whether // the number is Peterson number or not // To quickly find factorial of digits let fact = [ 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880 ]; // Function to check if a number is // Peterson or not function peterson(n) { let num = n, sum = 0; // stores the sum of factorials of // each digit of the number. while (n > 0) { let digit = n % 10; sum += fact[digit]; n = parseInt(n / 10); } // Condition check for a number to // be a Peterson Number return (sum == num); } // Driver code let n = 145; if (peterson(n)) document.write( "Yes" ); else document.write( "No" ); // This code is contributed by souravmahato348 </script> |
Yes
Time Complexity: log10(n)
Auxiliary Space: O(1)
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