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.

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
?>

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>

Output: 

Yes
No

Time Complexity: O(logn)

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

This article is contributed by Pramod Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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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”.

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))

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.