Given two positive integers B and N. The task is to find the number of trailing zeroes in b-ary (base B) representation of N! (factorial of N)
Examples:
Input: N = 5, B = 2 Output: 3 5! = 120 which is represented as 1111000 in base 2. Input: N = 6, B = 9 Output: 1
A naive solution is to find the factorial of the given number and convert it into given base B. Then, count the number of trailing zeroes but that would be a costly operation. Also, it will not be easy to find the factorial of large numbers and store it in integer.
Efficient Approach: Suppose, the base is 10 i.e., decimal then we’ll have to calculate the highest power of 10 that divides N! using Legendre’s formula. Thus, number B is represented as 10 when converted into base B. Let’s say base B = 13, then 13 in base 13 will be represented as 10, i.e., 1310 = 1013. Hence, problem reduces to finding the highest power of B in N!. (Largest power of k in n!)
Steps to solve the problem:
1. Function findPowerOfP takes two integer inputs N and p and returns an integer value count.
*Initialize a variable count to 0 and another variable r to p.
*While r is less than or equal to N, do the following steps:
*Calculate floor(N/r) and add it to count.
*Multiply r with p.
*Return count.
2. Function primeFactorsofB takes an integer input B and returns a vector of pairs of integers.
*Initialize an empty vector ans.
*Iterate from i=2 until B is not equal to 1, do the following:
*If B is divisible by i, initialize a variable count to 0 and do the following:
*While B is divisible by i, divide B by i and increment count.
*Add a pair of i and count to ans.
*Return ans.
3. Function largestPowerOfB takes two integer inputs N and B and returns an integer value.
*Initialize a vector of pairs vec and assign it the value returned by the function primeFactorsofB with input B.
*Initialize a variable ans to INT_MAX.
*Iterate from i=0 until i is less than the size of vec, do the following:
*Calculate the minimum of ans and the result of findPowerOfP with inputs N and vec[i].first divided by vec[i].second.
*Return ans.
Below is the implementation of the above approach.
// CPP program to find the number of trailing // zeroes in base B representation of N! #include <bits/stdc++.h> using namespace std;
// To find the power of a prime p in // factorial N int findPowerOfP( int N, int p)
{ int count = 0;
int r = p;
while (r <= N) {
// calculating floor(n/r)
// and adding to the count
count += (N / r);
// increasing the power of p
// from 1 to 2 to 3 and so on
r = r * p;
}
return count;
} // returns all the prime factors of k vector<pair< int , int > > primeFactorsofB( int B)
{ // vector to store all the prime factors
// along with their number of occurrence
// in factorization of B
vector<pair< int , int > > ans;
for ( int i = 2; B != 1; i++) {
if (B % i == 0) {
int count = 0;
while (B % i == 0) {
B = B / i;
count++;
}
ans.push_back(make_pair(i, count));
}
}
return ans;
} // Returns largest power of B that // divides N! int largestPowerOfB( int N, int B)
{ vector<pair< int , int > > vec;
vec = primeFactorsofB(B);
int ans = INT_MAX;
for ( int i = 0; i < vec.size(); i++)
// calculating minimum power of all
// the prime factors of B
ans = min(ans, findPowerOfP(N,
vec[i].first)
/ vec[i].second);
return ans;
} // Driver code int main()
{ cout << largestPowerOfB(5, 2) << endl;
cout << largestPowerOfB(6, 9) << endl;
return 0;
} |
// Java program to find the number of trailing // zeroes in base B representation of N! import java.util.*;
class GFG
{ static class pair
{ int first, second;
public pair( int first, int second)
{
this .first = first;
this .second = second;
}
} // To find the power of a prime p in // factorial N static int findPowerOfP( int N, int p)
{ int count = 0 ;
int r = p;
while (r <= N)
{
// calculating floor(n/r)
// and adding to the count
count += (N / r);
// increasing the power of p
// from 1 to 2 to 3 and so on
r = r * p;
}
return count;
} // returns all the prime factors of k static Vector<pair> primeFactorsofB( int B)
{ // vector to store all the prime factors
// along with their number of occurrence
// in factorization of B
Vector<pair> ans = new Vector<pair>();
for ( int i = 2 ; B != 1 ; i++)
{
if (B % i == 0 )
{
int count = 0 ;
while (B % i == 0 )
{
B = B / i;
count++;
}
ans.add( new pair(i, count));
}
}
return ans;
} // Returns largest power of B that // divides N! static int largestPowerOfB( int N, int B)
{ Vector<pair> vec = new Vector<pair>();
vec = primeFactorsofB(B);
int ans = Integer.MAX_VALUE;
for ( int i = 0 ; i < vec.size(); i++)
// calculating minimum power of all
// the prime factors of B
ans = Math.min(ans, findPowerOfP(
N, vec.get(i).first) /
vec.get(i).second);
return ans;
} // Driver code public static void main(String[] args)
{ System.out.println(largestPowerOfB( 5 , 2 ));
System.out.println(largestPowerOfB( 6 , 9 ));
} } // This code is contributed by Princi Singh |
# Python 3 program to find the number of # trailing zeroes in base B representation of N! import sys
# To find the power of a prime # p in factorial N def findPowerOfP(N, p):
count = 0
r = p
while (r < = N):
# calculating floor(n/r)
# and adding to the count
count + = int (N / r)
# increasing the power of p
# from 1 to 2 to 3 and so on
r = r * p
return count
# returns all the prime factors of k def primeFactorsofB(B):
# vector to store all the prime factors
# along with their number of occurrence
# in factorization of B'
ans = []
i = 2
while (B! = 1 ):
if (B % i = = 0 ):
count = 0
while (B % i = = 0 ):
B = int (B / i)
count + = 1
ans.append((i, count))
i + = 1
return ans
# Returns largest power of B that # divides N! def largestPowerOfB(N, B):
vec = []
vec = primeFactorsofB(B)
ans = sys.maxsize
# calculating minimum power of all
# the prime factors of B
ans = min (ans, int (findPowerOfP(N, vec[ 0 ][ 0 ]) /
vec[ 0 ][ 1 ]))
return ans
# Driver code if __name__ = = '__main__' :
print (largestPowerOfB( 5 , 2 ))
print (largestPowerOfB( 6 , 9 ))
# This code is contributed by # Surendra_Gangwar |
// C# program to find the number of trailing // zeroes in base B representation of N! using System;
using System.Collections.Generic;
class GFG
{ public class pair
{ public int first, second;
public pair( int first, int second)
{
this .first = first;
this .second = second;
}
} // To find the power of a prime p in // factorial N static int findPowerOfP( int N, int p)
{ int count = 0;
int r = p;
while (r <= N)
{
// calculating floor(n/r)
// and adding to the count
count += (N / r);
// increasing the power of p
// from 1 to 2 to 3 and so on
r = r * p;
}
return count;
} // returns all the prime factors of k static List<pair> primeFactorsofB( int B)
{ // vector to store all the prime factors
// along with their number of occurrence
// in factorization of B
List<pair> ans = new List<pair>();
for ( int i = 2; B != 1; i++)
{
if (B % i == 0)
{
int count = 0;
while (B % i == 0)
{
B = B / i;
count++;
}
ans.Add( new pair(i, count));
}
}
return ans;
} // Returns largest power of B that // divides N! static int largestPowerOfB( int N, int B)
{ List<pair> vec = new List<pair>();
vec = primeFactorsofB(B);
int ans = int .MaxValue;
for ( int i = 0; i < vec.Count; i++)
// calculating minimum power of all
// the prime factors of B
ans = Math.Min(ans, findPowerOfP(
N, vec[i].first) /
vec[i].second);
return ans;
} // Driver code public static void Main(String[] args)
{ Console.WriteLine(largestPowerOfB(5, 2));
Console.WriteLine(largestPowerOfB(6, 9));
} } // This code is contributed by 29AjayKumar |
<script> // JavaScript program to find the number of trailing // zeroes in base B representation of N! // To find the power of a prime p in // factorial N function findPowerOfP(N, p)
{ var count = 0;
var r = p;
while (r <= N) {
// calculating floor(n/r)
// and adding to the count
count += (N / r);
// increasing the power of p
// from 1 to 2 to 3 and so on
r = r * p;
}
return count;
} // returns all the prime factors of k function primeFactorsofB(B)
{ // vector to store all the prime factors
// along with their number of occurrence
// in factorization of B
var ans = [];
for ( var i = 2; B != 1; i++) {
if (B % i == 0) {
var count = 0;
while (B % i == 0) {
B = B / i;
count++;
}
ans.push([i, count]);
}
}
return ans;
} // Returns largest power of B that // divides N! function largestPowerOfB(N, B)
{ var vec =[];
vec = primeFactorsofB(B);
var ans = Number.MAX_VALUE;
for ( var i = 0; i < vec.length; i++)
// calculating minimum power of all
// the prime factors of B
ans = Math.min(ans, Math.floor(findPowerOfP(N,
vec[i][0]) / vec[i][1]));
return ans;
} // Driver code document.write(largestPowerOfB(5, 2) + "<br>" );
document.write(largestPowerOfB(6, 9) + "<br>" );
// This code is contributed by ShubhamSingh10 </script> |
3 1
Time Complexity : O(logN * logB)
Space Complexity : O(logB)