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 <iostream>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 functionint main(){ cout << "Count is " << countFact(2, 720); return 0;} |
Java
// Java Program to count factorial// numbers in given rangeclass 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 rangedef 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 codeprint("Count is ", countFact(2, 720))# This code is contributed# by Anant Agarwal. |
C#
// C# Program to count factorial// numbers in given rangeusing 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
<?php// PHP Program to count factorial// numbers in given rangefunction 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++; } // Count factorial numbers // in range [low, high] $res = 0; while ($fact <= $high) { $res++; $fact = $fact * $x; $x++; } // Return the count return $res;}// Driver Codeecho "Count is " , countFact(2, 720);// This code is contributed by ajit ?> |
Javascript
<script>// Javascript Program to count factorial// numbers in given rangefunction countFact(low, high){ // Find the first factorial // number 'fact' greater // than or equal to 'low' let fact = 1; let x = 1; while (fact < low) { fact = fact * x; x++; } // Count factorial numbers // in range [low, high] let res = 0; while (fact <= high) { res++; fact = fact * x; x++; } // Return the count return res;}// Driver Codedocument.write("Count is " + countFact(2, 720));// This code is contributed by _saurabh_jaiswal</script> |
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).
This article is contributed by Shivam. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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