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Largest power of k in n! (factorial) where k may not be prime
  • Difficulty Level : Medium
  • Last Updated : 16 Dec, 2020

Given two numbers k and n, find the largest power of k that divides n! 
Constraints: 
 

 K > 1

Examples: 
 

Input : n = 7, k = 2
Output : 4
Explanation : 7! = 5040
The largest power of 2 that
divides 5040 is 24.

Input : n = 10, k = 9
Output :  2
The largest power of 9 that
divides 10! is 92.

 

We have discussed a solution in below post when k is always prime.
Legendre’s formula (Given p and n, find the largest x such that p^x divides n!)
Now to find the power of any non-prime number k in n!, we first find all the prime factors of the number k along with the count of number of their occurrences. Then for each prime factor, we count occurrences using Legendre’s formula which states that the largest possible power of a prime number p in n is ⌊n/p⌋ + ⌊n/(p2)⌋ + ⌊n/(p3)⌋ + ……
Over all the prime factors p of K, the one with the minimum value of findPowerOfK(n, p)/count will be our answer where count is number of occurrences of p in k.
 

C++




// CPP program to find the largest power
// of k that divides 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> > primeFactorsofK(int k)
{
    // vector to store all the prime factors
    // along with their number of occurrence
    // in factorization of k
    vector<pair<int, int> > ans;
 
    for (int i = 2; k != 1; i++) {
        if (k % i == 0) {
            int count = 0;
            while (k % i == 0) {
                k = k / i;
                count++;
            }
 
            ans.push_back(make_pair(i, count));
        }
    }
    return ans;
}
 
// Returns largest power of k that
// divides n!
int largestPowerOfK(int n, int k)
{
    vector<pair<int, int> > vec;
    vec = primeFactorsofK(k);
    int ans = INT_MAX;
    for (int i = 0; i < vec.size(); i++)
 
        // calculating minimum power of all
        // the prime factors of k
        ans = min(ans, findPowerOfP(n,
              vec[i].first) / vec[i].second);
 
    return ans;
}
 
// Driver code
int main()
{
    cout << largestPowerOfK(7, 2) << endl;
    cout << largestPowerOfK(10, 9) << endl;
    return 0;
}

Java




// JAVA program to find the largest power
// of k that divides 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 Math.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 > primeFactorsofK(int k)
{
    // vector to store all the prime factors
    // along with their number of occurrence
    // in factorization of k
    Vector<pair> ans = new Vector<pair>();
 
    for (int i = 2; k != 1; i++)
    {
        if (k % i == 0)
        {
            int count = 0;
            while (k % i == 0)
            {
                k = k / i;
                count++;
            }
 
            ans.add(new pair(i, count));
        }
    }
    return ans;
}
 
// Returns largest power of k that
// divides n!
static int largestPowerOfK(int n, int k)
{
    Vector<pair > vec = new Vector<pair>();
    vec = primeFactorsofK(k);
    int ans = Integer.MAX_VALUE;
    for (int i = 0; i < vec.size(); i++)
 
        // calculating minimum power of all
        // the prime factors of k
        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.print(largestPowerOfK(7, 2) +"\n");
    System.out.print(largestPowerOfK(10, 9) +"\n");
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 program to find the largest power
# of k that divides 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 += (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 primeFactorsofK(k) :
 
    # vector to store all the prime factors
    # along with their number of occurrence
    # in factorization of k
    ans = []
    i = 2
    while k != 1 :
        if k % i == 0 :
            count = 0
            while k % i == 0 :
                k = k // i
                count += 1
            ans.append([i , count])
        i += 1
 
    return ans
 
# Returns largest power of k that
# divides n!
def largestPowerOfK(n, k) :
 
    vec = primeFactorsofK(k)
    ans = sys.maxsize
    for i in range(len(vec)) :
 
        # calculating minimum power of all
        # the prime factors of k
        ans = min(ans, findPowerOfP(n, vec[i][0]) // vec[i][1])
 
    return ans
 
print(largestPowerOfK(7, 2))
print(largestPowerOfK(10, 9))
 
# This code is contributed by divyesh072019

C#




// C# program to find the largest power
// of k that divides n!
using System;
using System.Collections.Generic;
 
class GFG
{
     
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 Math.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 > primeFactorsofK(int k)
{
    // vector to store all the prime factors
    // along with their number of occurrence
    // in factorization of k
    List<pair> ans = new List<pair>();
 
    for (int i = 2; k != 1; i++)
    {
        if (k % i == 0)
        {
            int count = 0;
            while (k % i == 0)
            {
                k = k / i;
                count++;
            }
 
            ans.Add(new pair(i, count));
        }
    }
    return ans;
}
 
// Returns largest power of k that
// divides n!
static int largestPowerOfK(int n, int k)
{
    List<pair > vec = new List<pair>();
    vec = primeFactorsofK(k);
    int ans = int.MaxValue;
    for (int i = 0; i < vec.Count; i++)
 
        // calculating minimum power of all
        // the prime factors of k
        ans = Math.Min(ans, findPowerOfP(n,
            vec[i].first) / vec[i].second);
 
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    Console.Write(largestPowerOfK(7, 2) +"\n");
    Console.Write(largestPowerOfK(10, 9) +"\n");
}
}
 
// This code is contributed by 29AjayKumar

Output: 
 

4
2

This article is contributed by ShivamKD. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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