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Largest power of k in n! (factorial) where k may not be prime

  • Difficulty Level : Medium
  • Last Updated : 08 Jul, 2021

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

 K > 1

Examples: 
 

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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

Javascript




<script>
 
// JavaScript program to find the largest power
// of k that divides n!
 
class pair
{
    constructor(first,second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// To find the power of a prime p in
// factorial N
function findPowerOfP(n,p)
{
    let count = 0;
    let r = p;
    while (r <= n)
    {
  
        // calculating Math.floor(n/r)
        // and adding to the count
        count += Math.floor(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 primeFactorsofK(k)
{
    // vector to store all the prime factors
    // along with their number of occurrence
    // in factorization of k
    let ans = [];
  
    for (let i = 2; k != 1; i++)
    {
        if (k % i == 0)
        {
            let count = 0;
            while (k % i == 0)
            {
                k = Math.floor(k / i);
                count++;
            }
  
            ans.push(new pair(i, count));
        }
    }
    return ans;
}
 
// Returns largest power of k that
// divides n!
function largestPowerOfK(n,k)
{
    let vec = [];
    vec = primeFactorsofK(k);
    let ans = Number.MAX_VALUE;
    for (let i = 0; i < vec.length; 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
document.write(largestPowerOfK(7, 2) +"<br>");
document.write(largestPowerOfK(10, 9) +"<br>");
 
 
// This code is contributed by rag2127
 
</script>

Output:  

4
2

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