# 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```
Recommended Practice
```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 ` `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

 ``

## Javascript

 ``

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 ` `#include ` `#include ` `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))`

## 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.

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