# 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 = 145Output : YesSum of digit factorials = 1! + 4! + 5!                        = 1 + 24 + 120                        = 145Input :  n = 534Output : 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.`

## Javascript

 ``

## PHP

 ``

Output

```Yes
No

```

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

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.

#### 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))`

## 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.

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next