Number of trailing zeroes in base B representation of N!
Last Updated :
04 Mar, 2023
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.
C++
#include <bits/stdc++.h>
using namespace std;
int findPowerOfP(int N, int p)
{
int count = 0;
int r = p;
while (r <= N) {
count += (N / r);
r = r * p;
}
return count;
}
vector<pair<int, int> > primeFactorsofB(int 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;
}
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++)
ans = min(ans, findPowerOfP(N,
vec[i].first)
/ vec[i].second);
return ans;
}
int main()
{
cout << largestPowerOfB(5, 2) << endl;
cout << largestPowerOfB(6, 9) << endl;
return 0;
}
|
Java
import java.util.*;
class GFG
{
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
static int findPowerOfP(int N, int p)
{
int count = 0;
int r = p;
while (r <= N)
{
count += (N / r);
r = r * p;
}
return count;
}
static Vector<pair> primeFactorsofB(int 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;
}
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++)
ans = Math.min(ans, findPowerOfP(
N, vec.get(i).first) /
vec.get(i).second);
return ans;
}
public static void main(String[] args)
{
System.out.println(largestPowerOfB(5, 2));
System.out.println(largestPowerOfB(6, 9));
}
}
|
Python3
import sys
def findPowerOfP(N, p):
count = 0
r = p
while (r <= N):
count += int(N / r)
r = r * p
return count
def primeFactorsofB(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
def largestPowerOfB(N, B):
vec = []
vec = primeFactorsofB(B)
ans = sys.maxsize
ans = min(ans, int(findPowerOfP(N, vec[0][0]) /
vec[0][1]))
return ans
if __name__ == '__main__':
print(largestPowerOfB(5, 2))
print(largestPowerOfB(6, 9))
|
C#
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;
}
}
static int findPowerOfP(int N, int p)
{
int count = 0;
int r = p;
while (r <= N)
{
count += (N / r);
r = r * p;
}
return count;
}
static List<pair> primeFactorsofB(int 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;
}
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++)
ans = Math.Min(ans, findPowerOfP(
N, vec[i].first) /
vec[i].second);
return ans;
}
public static void Main(String[] args)
{
Console.WriteLine(largestPowerOfB(5, 2));
Console.WriteLine(largestPowerOfB(6, 9));
}
}
|
Javascript
<script>
function findPowerOfP(N, p)
{
var count = 0;
var r = p;
while (r <= N) {
count += (N / r);
r = r * p;
}
return count;
}
function primeFactorsofB(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;
}
function largestPowerOfB(N, B)
{
var vec =[];
vec = primeFactorsofB(B);
var ans = Number.MAX_VALUE;
for (var i = 0; i < vec.length; i++)
ans = Math.min(ans, Math.floor(findPowerOfP(N,
vec[i][0]) / vec[i][1]));
return ans;
}
document.write(largestPowerOfB(5, 2) + "<br>");
document.write(largestPowerOfB(6, 9) + "<br>");
</script>
|
Time Complexity : O(logN * logB)
Space Complexity : O(logB)
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