Check whether N is a Factorion or not
Given an integer N and the task is to check whether N is a Factorion or not. A Factorion is a number which is equal to the sum of the factorials of its digits.
Examples:
Input: N = 40585
Output: Yes
4! + 0! + 5! + 8! + 5! = 40585
Input: N = 234
Output: No
2! + 3! + 4! = 32
Approach: Create an array fact[] of size 10 to store the factorials of all possible digits where fact[i] stores i!. Now for all the digits of the given number find the sum of factorials of the digits using the fact[] array computed earlier. If the sum if equal to the given number then the number is a Factorion else it is not.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define MAX 10// Function that returns true// if n is a Factorionbool isFactorion(int n){ // fact[i] will store i! int fact[MAX]; fact[0] = 1; for (int i = 1; i < MAX; i++) fact[i] = i * fact[i - 1]; // A copy of the given integer int org = n; // To store the sum of factorials // of the digits of n int sum = 0; while (n > 0) { // Get the last digit int d = n % 10; // Add the factorial of the current // digit to the sum sum += fact[d]; // Remove the last digit n /= 10; } if (sum == org) return true; return false;}// Driver codeint main(){ int n = 40585; if (isFactorion(n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the above approachclass GFG { static int MAX = 10; // Function that returns true // if n is a Factorion static boolean isFactorion(int n) { // fact[i] will store i! int fact[] = new int[MAX]; fact[0] = 1; for (int i = 1; i < MAX; i++) fact[i] = i * fact[i - 1]; // A copy of the given integer int org = n; // To store the sum of factorials // of the digits of n int sum = 0; while (n > 0) { // Get the last digit int d = n % 10; // Add the factorial of the current // digit to the sum sum += fact[d]; // Remove the last digit n /= 10; } if (sum == org) return true; return false; } // Driver code public static void main (String[] args) { int n = 40585; if (isFactorion(n)) System.out.println("Yes"); else System.out.println("No"); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach MAX = 10# Function that returns true # if n is a Factorion def isFactorion(n) : # fact[i] will store i! fact = [0] * MAX fact[0] = 1 for i in range(1, MAX) : fact[i] = i * fact[i - 1] # A copy of the given integer org = n # To store the sum of factorials # of the digits of n sum = 0 while (n > 0) : # Get the last digit d = n % 10 # Add the factorial of the current # digit to the sum sum += fact[d] # Remove the last digit n = n // 10 if (sum == org): return True return False# Driver code n = 40585if (isFactorion(n)): print("Yes") else: print("No") # This code is contributed by# divyamohan123 |
C#
// C# implementation of the above approachusing System;class GFG { static int MAX = 10; // Function that returns true // if n is a Factorion static bool isFactorion(int n) { // fact[i] will store i! int [] fact = new int[MAX]; fact[0] = 1; for (int i = 1; i < MAX; i++) fact[i] = i * fact[i - 1]; // A copy of the given integer int org = n; // To store the sum of factorials // of the digits of n int sum = 0; while (n > 0) { // Get the last digit int d = n % 10; // Add the factorial of the current // digit to the sum sum += fact[d]; // Remove the last digit n /= 10; } if (sum == org) return true; return false; } // Driver code public static void Main (String[] args) { int n = 40585; if (isFactorion(n)) Console.WriteLine("Yes"); else Console.WriteLine("No"); } }// This code is contributed by Mohit kumar |
Javascript
<script>// Javascript implementation of the approachconst MAX = 10;// Function that returns true// if n is a Factorionfunction isFactorion(n){ // fact[i] will store i! let fact = new Array(MAX); fact[0] = 1; for (let i = 1; i < MAX; i++) fact[i] = i * fact[i - 1]; // A copy of the given integer let org = n; // To store the sum of factorials // of the digits of n let sum = 0; while (n > 0) { // Get the last digit let d = n % 10; // Add the factorial of the current // digit to the sum sum += fact[d]; // Remove the last digit n = parseInt(n / 10); } if (sum == org) return true; return false;}// Driver code let n = 40585; if (isFactorion(n)) document.write("Yes"); else document.write("No");</script> |
Output:
Yes
Time Complexity: O(log10n)
Auxiliary Space: O(MAX)
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