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# Program to check Strong Number

• Difficulty Level : Easy
• Last Updated : 24 Mar, 2023

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;`   `// Fills factorials of digits from 0 to 9.` `void` `preCompute()` `{` `    ``f = f = 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``;`   `    ``// Fills factorials of digits from 0 to 9.` `    ``static` `void` `preCompute()` `    ``{` `        ``f = f = 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.

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.

#### 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.

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