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# Count factorial numbers in a given range

• Difficulty Level : Easy
• Last Updated : 03 Jan, 2023

A number F is a factorial number if there exists some integer I >= 0 such that F = I! (that is, F is factorial of I). Examples of factorial numbers are 1, 2, 6, 24, 120, ….
Write a program that takes as input two long integers ‘low’ and ‘high’ where 0 < low < high and finds count of factorial numbers in the closed interval [low, high].
Examples :

```Input: 0 1
Output: 1 //Reason: Only factorial number is 1

Input: 12 122
Output: 2 // Reason: factorial numbers are 24, 120

Input: 2 720
Output: 5 // Factorial numbers are: 2, 6, 24, 120, 720 ```

1) Find the first factorial that is greater than or equal to low. Let this factorial be x! (factorial of x) and value of this factorial be ‘fact’
2) Keep incrementing x, and keep updating ‘fact’ while fact is smaller than or equal to high. Count the number of times, this loop runs.
3) Return the count computed in step 2.
Below is implementation of above algorithm. Thanks to Kartik for suggesting below solution.

## C++

 `// C++ Program to count factorial numbers in given range``#include ``using` `namespace` `std;` `int` `countFact(``int` `low, ``int` `high)``{``    ``// Find the first factorial number 'fact' greater than or``    ``// equal to 'low'``    ``int` `fact = 1, x = 1;``    ``while` `(fact < low)``    ``{``        ``fact = fact*x;``        ``x++;``    ``}` `    ``// Count factorial numbers in range [low, high]``    ``int` `res = 0;``    ``while` `(fact <= high)``    ``{``        ``res++;``        ``fact = fact*x;``        ``x++;``    ``}` `    ``// Return the count``    ``return` `res;``}` `// Driver program to test above function``int` `main()``{``    ``cout << ``"Count is "` `<< countFact(2, 720);``    ``return` `0;``}`

## Java

 `// Java Program to count factorial``// numbers in given range` `class` `GFG``{``    ``static` `int` `countFact(``int` `low, ``int` `high)``    ``{``        ``// Find the first factorial number``        ``// 'fact' greater than or equal to 'low'``        ``int` `fact = ``1``, x = ``1``;``        ``while` `(fact < low)``        ``{``            ``fact = fact * x;``            ``x++;``        ``}``    ` `        ``// Count factorial numbers``        ``// in range [low, high]``        ``int` `res = ``0``;``        ``while` `(fact <= high)``        ``{``            ``res++;``            ``fact = fact * x;``            ``x++;``        ``}``    ` `        ``// Return the count``        ``return` `res;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``System.out.print(``"Count is "``                         ``+ countFact(``2``, ``720``));``    ``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python3 Program to count factorial``# numbers in given range` `def` `countFact(low,high):` `    ``# Find the first factorial number``    ``# 'fact' greater than or``    ``# equal to 'low'``    ``fact ``=` `1``    ``x ``=` `1``    ``while` `(fact < low):``    ` `        ``fact ``=` `fact ``*` `x``        ``x ``+``=` `1``    ` ` ` `    ``# Count factorial numbers``    ``# in range [low, high]``    ``res ``=` `0``    ``while` `(fact <``=` `high):``    ` `        ``res ``+``=` `1``        ``fact ``=` `fact ``*` `x``        ``x ``+``=` `1``    ` ` ` `    ``# Return the count``    ``return` `res` `# Driver code` `print``(``"Count is "``, countFact(``2``, ``720``))` `# This code is contributed``# by Anant Agarwal.`

## C#

 `// C# Program to count factorial``// numbers in given range``using` `System;` `public` `class` `GFG``{``    ` `    ``// Function to count factorial``    ``static` `int` `countFact(``int` `low, ``int` `high)``    ``{``        ` `        ``// Find the first factorial number numbers``        ``// 'fact' greater than or equal to 'low'``        ``int` `fact = 1, x = 1;``        ``while` `(fact < low)``        ``{``            ``fact = fact * x;``            ``x++;``        ``}``    ` `        ``// Count factorial numbers``        ``// in range [low, high]``        ``int` `res = 0;``        ``while` `(fact <= high)``        ``{``            ``res++;``            ``fact = fact * x;``            ``x++;``        ``}``    ` `        ``// Return the count``        ``return` `res;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``Console.Write(``"Count is "` `+ countFact(2, 720));``    ``}``}` `// This code is contributed by Sam007`

## PHP

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## Javascript

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Output :

`Count is 5`

Time complexity: approximately equal to O(K) where K is the biggest divisor of high and is not equal to high.

Auxiliary space complexity:  O(1).