Smallest number with at least n trailing zeroes in factorial
Given a number n. The task is to find the smallest number whose factorial contains at least n trailing zeroes.
Examples :
Input : n = 1 Output : 5 1!, 2!, 3!, 4! does not contain trailing zero. 5! = 120, which contains one trailing zero. Input : n = 6 Output : 25
In the article for Count trailing zeroes in factorial of a number, we have discussed number of zeroes is equal to number of 5’s in prime factors of x!. We have discussed below formula to count number of 5’s.
Trailing 0s in x! = Count of 5s in prime factors of x! = floor(x/5) + floor(x/25) + floor(x/125) + ....
Let us take few examples to observe pattern
5! has 1 trailing zeroes [All numbers from 6 to 9 have 1 trailing zero] 10! has 2 trailing zeroes [All numbers from 11 to 14 have 2 trailing zeroes] 15! to 19! have 3 trailing zeroes 20! to 24! have 4 trailing zeroes 25! to 29! have 6 trailing zeroes
We can notice that, the maximum value whose factorial contain n trailing zeroes is 5*n.
So, to find minimum value whose factorial contains n trailing zeroes, use binary search on range from 0 to 5*n. And, find the smallest number whose factorial contains n trailing zeroes.
C++
// C++ program to find smallest number whose // factorial contains at least n trailing // zeroes. #include<bits/stdc++.h> using namespace std; // Return true if number's factorial contains // at least n trailing zero else false. bool check( int p, int n) { int temp = p, count = 0, f = 5; while (f <= temp) { count += temp/f; f = f*5; } return (count >= n); } // Return smallest number whose factorial // contains at least n trailing zeroes int findNum( int n) { // If n equal to 1, return 5. // since 5! = 120. if (n==1) return 5; // Initialising low and high for binary // search. int low = 0; int high = 5*n; // Binary Search. while (low <high) { int mid = (low + high) >> 1; // Checking if mid's factorial contains // n trailing zeroes. if (check(mid, n)) high = mid; else low = mid+1; } return low; } // driver code int main() { int n = 6; cout << findNum(n) << endl; return 0; } |
Java
// Java program to find smallest number whose // factorial contains at least n trailing // zeroes. class GFG { // Return true if number's factorial contains // at least n trailing zero else false. static boolean check( int p, int n) { int temp = p, count = 0 , f = 5 ; while (f <= temp) { count += temp / f; f = f * 5 ; } return (count >= n); } // Return smallest number whose factorial // contains at least n trailing zeroes static int findNum( int n) { // If n equal to 1, return 5. // since 5! = 120. if (n== 1 ) return 5 ; // Initialising low and high for binary // search. int low = 0 ; int high = 5 * n; // Binary Search. while (low < high) { int mid = (low + high) >> 1 ; // Checking if mid's factorial // contains n trailing zeroes. if (check(mid, n)) high = mid; else low = mid + 1 ; } return low; } // Driver code public static void main (String[] args) { int n = 6 ; System.out.println(findNum(n)); } } // This code is contributed by Anant Agarwal. |
Python3
# Python3 program to find smallest # number whose # factorial contains at least # n trailing zeroes # Return true if number's factorial contains # at least n trailing zero else false. def check(p,n): temp = p count = 0 f = 5 while (f < = temp): count + = temp / / f f = f * 5 return (count > = n) # Return smallest number whose factorial # contains at least n trailing zeroes def findNum(n): # If n equal to 1, return 5. # since 5! = 120. if (n = = 1 ): return 5 # Initializing low and high for binary # search. low = 0 high = 5 * n # Binary Search. while (low <high): mid = (low + high) >> 1 # Checking if mid's factorial contains # n trailing zeroes. if (check(mid, n)): high = mid else : low = mid + 1 return low # driver code n = 6 print (findNum(n)) # This code is contributed # by Anant Agarwal. |
C#
// C# program to find smallest number whose // factorial contains at least n trailing // zeroes. using System; class GFG { // Return true if number's factorial contains // at least n trailing zero else false. static bool check( int p, int n) { int temp = p, count = 0, f = 5; while (f <= temp) { count += temp / f; f = f * 5; } return (count >= n); } // Return smallest number whose factorial // contains at least n trailing zeroes static int findNum( int n) { // If n equal to 1, return 5. // since 5! = 120. if (n == 1) return 5; // Initialising low and high for binary // search. int low = 0; int high = 5 * n; // Binary Search. while (low < high) { int mid = (low + high) >> 1; // Checking if mid's factorial // contains n trailing zeroes. if (check(mid, n)) high = mid; else low = mid + 1; } return low; } // Driver code public static void Main () { int n = 6; Console.WriteLine(findNum(n)); } } // This code is contributed by vt_m. |
PHP
<?php // PHP program to find smallest // number whose factorial contains // at least n trailing zeroes. // Return true if number's // factorial contains at // least n trailing zero // else false. function check( $p , $n ) { $temp = $p ; $count = 0; $f = 5; while ( $f <= $temp ) { $count += $temp / $f ; $f = $f * 5; } return ( $count >= $n ); } // Return smallest number // whose factorial contains // at least n trailing zeroes function findNum( $n ) { // If n equal to 1, return 5. // since 5! = 120. if ( $n == 1) return 5; // Initialising low and high // for binary search. $low = 0; $high = 5 * $n ; // Binary Search. while ( $low < $high ) { $mid = ( $low + $high ) >> 1; // Checking if mid's factorial // contains n trailing zeroes. if (check( $mid , $n )) $high = $mid ; else $low = $mid + 1; } return $low ; } // Driver Code $n = 6; echo (findNum( $n )); // This code is contributed by Ajit. ?> |
Javascript
<script> // JavaScript program to find smallest number whose // factorial contains at least n trailing // zeroes. // Return true if number's factorial contains // at least n trailing zero else false. function check(p, n) { let temp = p, count = 0, f = 5; while (f <= temp) { count += Math.floor(temp/f); f = f*5; } return (count >= n); } // Return smallest number whose factorial // contains at least n trailing zeroes function findNum(n) { // If n equal to 1, return 5. // since 5! = 120. if (n==1) return 5; // Initialising low and high for binary // search. let low = 0; let high = 5*n; // Binary Search. while (low <high) { let mid = (low + high) >> 1; // Checking if mid's factorial contains // n trailing zeroes. if (check(mid, n)) high = mid; else low = mid+1; } return low; } // driver code let n = 6; document.write(findNum(n) + "<br>" ); // This code is contributed by Surbhi Tyagi </script> |
Output :
25
Time Complexity: O(log2N)
We take log2N in binary search and our check() function takes log5N time so the overall time complexity beacomes log2N * log5N which in a more general sense can be written as (logN)2 which can also be written as log2N.
Auxiliary Space: O(1)
As constant extra space is used.
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