Program to check Strong Number

Strong Numbers are the numbers whose sum of factorial of digits is equal to the original number. Given a number, check if it is a Strong Number or not.
Examples:

Input  : n = 145
Output : Yes
Sum of digit factorials = 1! + 4! + 5!
                        = 1 + 24 + 120
                        = 145
Input :  n = 534
Output : No

1) Initialize sum of factorials as 0. 
2) For every digit d, do following
   a) Add d! to sum of factorials.
3) If sum factorials is same as given 
   number, return true.
4) Else return false.

An optimization is to precompute factorials of all numbers from 0 to 10.

C++

// C++ program to check if a number is
// strong or not.
#include <bits/stdc++.h>
using namespace std;

int f[10];

// Fills factorials of digits from 0 to 9.
void preCompute()
{
    f[0] = f[1] = 1;
    for (int i = 2; i<10; ++i)
        f[i] = f[i-1] * i;
}

// Returns true if x is Strong
bool isStrong(int x)
{
    int factSum = 0;

    // Traverse through all digits of x.
    int temp = x;
    while (temp)
    {
        factSum += f[temp%10];
        temp /= 10;
    }

    return (factSum == x);
}

// Driver code
int main()
{
    preCompute();

    int x = 145;
    isStrong(x) ? cout << "Yes\n" : cout << "No\n";
    x = 534;
    isStrong(x) ? cout << "Yes\n" : cout << "No\n";
    return 0;
}

Java

// Java program to check if
// a number is Strong or not

class CheckStrong
{
    static int f[] = new int[10];

    // Fills factorials of digits from 0 to 9.
    static void preCompute()
    {
        f[0] = f[1] = 1;
        for (int i = 2; i<10; ++i)
            f[i] = f[i-1] * i;
    }

    // Returns true if x is Strong
    static boolean isStrong(int x)
    {
        int factSum = 0;

        // Traverse through all digits of x.
        int temp = x;
        while (temp>0)
        {
            factSum += f[temp%10];
            temp /= 10;
        }

        return (factSum == x);
    }

    // main function
    public static void main (String[] args)
    {
        // calling preCompute
        preCompute();

        // first pass
        int x = 145;
        if(isStrong(x))
        {
            System.out.println("Yes");
        }
        else
            System.out.println("No");

        // second pass
        x = 534;
        if(isStrong(x))
        {
            System.out.println("Yes");
        }
        else
            System.out.println("No");
    }
}

Python3

# Python program to check if a number is
# strong or not.

f = [None] * 10

# Fills factorials of digits from 0 to 9.
def preCompute() :
    f[0] = f[1] = 1;
    for i in range(2,10) :
        f[i] = f[i-1] * i

# Returns true if x is Strong
def isStrong(x) :

    factSum = 0
    # Traverse through all digits of x.
    temp = x
    while (temp) :
        factSum = factSum + f[temp % 10]
        temp = temp // 10

    return (factSum == x)

# Driver code
preCompute()
x = 145
if(isStrong(x) ) :
    print ("Yes")
else :
    print ("No")
x = 534
if(isStrong(x)) :
    print ("Yes")
else:
    print ("No")

# This code is contributed by Nikita Tiwari.

C#

// C# program to check if
// a number is Strong or not
using System;

class CheckStrong
{
    static int []f = new int[10];

    // Fills factorials of digits from 0 to 9.
    static void preCompute()
    {
        f[0] = f[1] = 1;
        for (int i = 2; i < 10; ++i)
            f[i] = f[i - 1] * i;
    }

    // Returns true if x is Strong
    static bool isStrong(int x)
    {
        int factSum = 0;

        // Traverse through all digits of x.
        int temp = x;
        while (temp > 0)
        {
            factSum += f[temp % 10];
            temp /= 10;
        }

        return (factSum == x);
    }

    // Driver Code
    public static void Main ()
    {
        // calling preCompute
        preCompute();

        // first pass
        int x = 145;
        if(isStrong(x))
        {
            Console.WriteLine("Yes");
        }
        else
            Console.WriteLine("No");

        // second pass
        x = 534;
        if(isStrong(x))
        {
            Console.WriteLine("Yes");
        }
        else
            Console.WriteLine("No");
    }
}

// This code is contributed by Nitin Mittal.

Javascript

<script>

// Javascript program to check if a number is
// strong or not.


let f = new Array(10);

// Fills factorials of digits from 0 to 9.
function preCompute()
{
    f[0] = f[1] = 1;
    for (let i = 2; i<10; ++i)
        f[i] = f[i-1] * i;
}

// Returns true if x is Strong
function isStrong(x)
{
    let factSum = 0;

    // Traverse through all digits of x.
    let temp = x;
    while (temp)
    {
        factSum += f[temp%10];
        temp = Math.floor(temp/10);
    }

    return (factSum == x);
}

// Driver code

    preCompute();

    let x = 145;
    isStrong(x) ? document.write("Yes" + "<br>") :
    document.write("No" + "<br>");
    x = 534;
    isStrong(x) ? document.write("Yes" + "<br>") :
    document.write("No" + "<br>");


//This code is contributed by Mayank Tyagi

</script>

PHP

<?php
// PHP program to check if a number
// is strong or not.

$f[10] = array();

// Fills factorials of digits
// from 0 to 9.
function preCompute()
{
    global $f;
    $f[0] = $f[1] = 1;
    for ($i = 2; $i < 10; ++$i)
        $f[$i] = $f[$i - 1] * $i;
}

// Returns true if x is Strong
function isStrong($x)
{
    global $f;
    $factSum = 0;

    // Traverse through all digits of x.
    $temp = $x;
    while ($temp)
    {
        $factSum += $f[$temp % 10];
        $temp = (int)$temp / 10;
    }

    return ($factSum == $x);
}

// Driver code
preCompute();
$x = 145;

if(isStrong(!$x))
    echo "Yes\n";
else
    echo "No\n";
$x = 534;
if(isStrong($x))
    echo "Yes\n";
else
    echo "No\n";

// This code is contributed by jit_t
?>

Output

Yes
No

Time Complexity: O(logn)
Auxiliary Space: O(1), since constant extra space has been taken.

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Approach#2: Using Iterative Method 

This approach converts the input number into a list of its digits. It then computes the factorial sum of each digit and adds them up. If the final sum is equal to the input number, it returns “Yes”, otherwise it returns “No”.

Algorithm

1. Define a function is_strong(n) that takes a number n as input.
2. Convert the number to a string and get its digits.
3. Calculate the factorial of each digit using an iterative method.
4. Sum the factorials of all digits.
5. If the sum is equal to the number n, return “Yes”, otherwise return “No”.

C++

// C++ code addition

#include <iostream>
#include <vector>
#include <string>
using namespace std;

string is_strong(int n) {
  // Convert the number to an array of digits
    vector<int> digits;
    int temp = n;
  // Loop through each digit
    while (temp != 0) {
        digits.insert(digits.begin(), temp % 10);
        temp /= 10;
    }
    int factorial_sum = 0;
  // Calculate the factorial of the digit
    for (int d : digits) {
        int f = 1;
        for (int i = 1; i <= d; i++) {
            f *= i;
        }
      // Add the factorial to the sum
        factorial_sum += f;
    }
   // Check if the sum of factorials is equal to the original number
    if (factorial_sum == n) {
        return "Yes";
    } else {
        return "No";
    }
}

int main() {
    int n = 145;
    cout << is_strong(n) << endl;
    return 0;
}

// The code is contributed by Arushi Goel.

Python3

def is_strong(n):
    digits = [int(d) for d in str(n)]
    factorial_sum = 0
    for d in digits:
        f = 1
        for i in range(1, d+1):
            f *= i
        factorial_sum += f
    if factorial_sum == n:
        return "Yes"
    else:
        return "No"
n=145
print(is_strong(n))

C#

using System;

class Program
{
    static string IsStrong(int n)
    {
        // Convert the number to an array of digits
        char[] digits = n.ToString().ToCharArray();
        int factorialSum = 0;

        // Loop through each digit
        foreach (char digit in digits)
        {
            int d = int.Parse(digit.ToString());
            int f = 1;

            // Calculate the factorial of the digit
            for (int i = 1; i <= d; i++)
            {
                f *= i;
            }

            // Add the factorial to the sum
            factorialSum += f;
        }

        // Check if the sum of factorials is equal to the original number
        return (factorialSum == n) ? "Yes" : "No";
    }

    static void Main()
    {
        int n = 145;
        Console.WriteLine(IsStrong(n));
    }
}

Javascript

// Function to check if a number is a strong number
function is_strong(n) {
    // Convert the number to an array of digits
    let digits = Array.from(String(n), Number);
    let factorial_sum = 0;
    // Loop through each digit
    for (let d of digits) {
        let f = 1;
        // Calculate the factorial of the digit
        for (let i = 1; i <= d; i++) {
            f *= i;
        }
        // Add the factorial to the sum
        factorial_sum += f;
    }
    // Check if the sum of factorials is equal to the original number
    if (factorial_sum == n) {
        return "Yes";
    } else {
        return "No";
    }
}

let n = 145;
console.log(is_strong(n));

Output

Yes

Time Complexity: O(kn^2), where k is the number of digits in the number and n is the maximum value of a digit
Auxiliary Space: O(k), where k is the number of digits in the number.