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Finding power of prime number p in n!

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Given a number ‘n’ and a prime number ‘p’. We need to find out the power of ‘p’ in the prime factorization of n!
Examples: 
 

Input  : n = 4, p = 2
Output : 3
         Power of 2 in the prime factorization
         of 2 in 4! = 24 is 3

Input  : n = 24, p = 2
Output : 22


 


Naive approach 
The naive approach is to find the power of p in each number from 1 to n and add them. Because we know that during multiplication power is added.
 

C++

// C++ implementation of finding
// power of p in n!
#include <bits/stdc++.h>
 
using namespace std;
 
// Returns power of p in n!
int PowerOFPINnfactorial(int n, int p)
{
    // initializing answer
    int ans = 0;
 
    // initializing
    int temp = p;
 
    // loop until temp<=n
    while (temp <= n) {
 
        // add number of numbers divisible by temp
        ans += n / temp;
 
        // each time multiply temp by p
        temp = temp * p;
    }
    return ans;
}
 
// Driver function
int main()
{
    cout << PowerOFPINnfactorial(4, 2) << endl;
    return 0;
}

                    

Java

// Java implementation of naive approach
 
public class GFG
{
    // Method to calculate the power of prime number p in n!
    static int PowerOFPINnfactorial(int n, int p)
    {
        // initializing answer
        int ans = 0;
      
        // finding power of p from 1 to n
        for (int i = 1; i <= n; i++) {
      
            // variable to note the power of p in i
            int count = 0, temp = i;
      
            // loop until temp is equal to i
            while (temp % p == 0) {
                count++;
                temp = temp / p;
            }
      
            // adding count to i
            ans += count;
        }
        return ans;
    }
     
    // Driver Method
    public static void main(String[] args)
    {
        System.out.println(PowerOFPINnfactorial(4, 2));
    }
}

                    

Python3

# Python3 implementation of
# finding power of p in n!
 
# Returns power of p in n!
def PowerOFPINnfactorial(n, p):
 
    # initializing answer
    ans = 0;
 
    # initializing
    temp = p;
 
    # loop until temp<=n
    while (temp <= n):
 
        # add number of numbers
        # divisible by n
        ans += n / temp;
 
        # each time multiply
        # temp by p
        temp = temp * p;
         
    return ans;
 
# Driver Code
print(PowerOFPINnfactorial(4, 2));
 
# This code is contributed by
# mits

                    

C#

// C# implementation of naive approach
using System;
 
public class GFG
{
    // Method to calculate power
    // of prime number p in n!
    static int PowerOFPINnfactorial(int n, int p)
    {
        // initializing answer
        int ans = 0;
     
        // finding power of p from 1 to n
        for (int i = 1; i <= n; i++) {
     
            // variable to note the power of p in i
            int count = 0, temp = i;
     
            // loop until temp is equal to i
            while (temp % p == 0) {
                count++;
                temp = temp / p;
            }
     
            // adding count to i
            ans += count;
        }
        return ans;
    }
     
    // Driver Code
    public static void Main(String []args)
    {
        Console.WriteLine(PowerOFPINnfactorial(4, 2));
    }
}
 
// This code is contributed by vt_m.

                    

PHP

<?php
// PHP implementation of
// finding power of p in n!
  
// Returns power of p in n!
function PowerOFPINnfactorial($n, $p)
{
    // initializing answer
    $ans = 0;
  
    // initializing
    $temp = $p;
  
    // loop until temp<=n
    while ($temp <= $n)
    {
  
        // add number of numbers
        // divisible by n
        $ans += $n / $temp;
  
        // each time multiply
        // temp by p
        $temp = $temp * $p;
    }
    return $ans;
}
  
// Driver Code
echo PowerOFPINnfactorial(4, 2) . "\n";
  
// This code is contributed by
// Akanksha Rai(Abby_akku)
?>

                    

Javascript

<script>
 
// Javascript implementation of
// finding power of p in n!
  
// Returns power of p in n!
function PowerOFPINnfactorial(n, p)
{
    // initializing answer
    let ans = 0;
  
    // initializing
    let temp = p;
  
    // loop until temp<=n
    while (temp <= n)
    {
  
        // add number of numbers
        // divisible by n
        ans += n / temp;
  
        // each time multiply
        // temp by p
        temp = temp * p;
    }
    return ans;
}
  
// Driver Code
document.write(PowerOFPINnfactorial(4, 2));
  
// This code is contributed by _saurabh_jaiswal
 
</script>

                    

Kotlin

//function to find the power of p in n! in Kotlin
fun PowerOFPINnfactorial(n: Int, p: Int)
{
    // initializing answer
    var ans = 0;
 
    // initializing
    var temp = p;
 
    // loop until temp<=n
    while(temp<=n)
    {
        // add number of numbers divisible by temp
        ans+=n/temp;
         
        // each time multiply temp by p
        temp*=p;
    }
 
    println(ans)
}
 
//Driver Code
fun main(args: Array<String>)
{
    val n = 4
    val p = 2
    PowerOFPINnfactorial(n,p)
}

                    

Output: 

3

Time Complexity: O(logpn) 
Auxiliary Space: O(1)


Efficient Approach 
Before discussing efficient approach lets discuss a question given a two numbers n, m how many numbers are there from 1 to n that are divisible by m the answer is floor(n/m). 
Now coming back to our original question to find the power of p in n! what we do is count the number of numbers from 1 to n that are divisible by p then by p^2       then by p^3       till p^k       > n and add them. This will be our required answer. 

   Powerofp(n!) = floor(n/p) + floor(n/p^2) + floor(n/p^3)........ 


Below is the implementation of the above steps.
 

C++

// C++ implementation of finding power of p in n!
#include <bits/stdc++.h>
 
using namespace std;
 
// Returns power of p in n!
int PowerOFPINnfactorial(int n, int p)
{
    // initializing answer
    int ans = 0;
 
    // initializing
    int temp = p;
 
    // loop until temp<=n
    while (temp <= n) {
 
        // add number of numbers divisible by temp
        ans += n / temp;
 
        // each time multiply temp by p
        temp = temp * p;
    }
    return ans;
}
 
// Driver function
int main()
{
    cout << PowerOFPINnfactorial(4, 2) << endl;
    return 0;
}

                    

Java

// Java implementation of finding power of p in n!
 
public class GFG
{
    // Method to calculate the power of prime number p in n!
    static int PowerOFPINnfactorial(int n, int p)
    {
        // initializing answer
        int ans = 0;
      
        // initializing
        int temp = p;
      
        // loop until temp<=n
        while (temp <= n) {
      
            // add number of numbers divisible by n
            ans += n / temp;
      
            // each time multiply temp by p
            temp = temp * p;
        }
        return ans;
    }
     
    // Driver Method
    public static void main(String[] args)
    {
        System.out.println(PowerOFPINnfactorial(4, 2));
    }
}

                    

Python3

# Python3 implementation of
# finding power of p in n!
 
# Returns power of p in n!
def PowerOFPINnfactorial(n, p):
 
    # initializing answer
    ans = 0
 
    # initializing
    temp = p
 
    # loop until temp<=n
    while (temp <= n) :
 
        # add number of numbers
        # divisible by n
        ans += n / temp
 
        # each time multiply
        # temp by p
        temp = temp * p
     
    return int(ans)
 
# Driver Code
print(PowerOFPINnfactorial(4, 2))
 
# This code is contributed
# by Smitha

                    

C#

// C# implementation of finding
// power of p in n!
using System;
 
public class GFG
{
 
    // Method to calculate power
    // of prime number p in n!
    static int PowerOFPINnfactorial(int n, int p)
    {
        // initializing answer
        int ans = 0;
     
        // initializing
        int temp = p;
     
        // loop until temp <= n
        while (temp <= n) {
     
            // add number of numbers divisible by n
            ans += n / temp;
     
            // each time multiply temp by p
            temp = temp * p;
        }
        return ans;
    }
     
    // Driver Code
    public static void Main(String []args)
    {
        Console.WriteLine(PowerOFPINnfactorial(4, 2));
    }
}
 
// This code is contributed by vt_m.

                    

PHP

<?php
// PHP implementation of
// finding power of p in n!
 
// Returns power of p in n!
function PowerOFPINnfactorial($n, $p)
{
    // initializing answer
    $ans = 0;
 
    // initializing
    $temp = $p;
 
    // loop until temp<=n
    while ($temp <= $n)
    {
 
        // add number of numbers divisible by n
        $ans += $n / $temp;
 
        // each time multiply temp by p
        $temp = $temp * $p;
    }
    return $ans;
}
 
// Driver function
echo PowerOFPINnfactorial(4, 2) . "\n";
 
// This code is contributed
// by Akanksha Rai(Abby_akku)
?>

                    

Javascript

<script>
 
// Javascript implementation of
// finding power of p in n!
 
// Returns power of p in n!
function PowerOFPINnfactorial(n, p)
{
    // initializing answer
    let ans = 0;
 
    // initializing
    let temp = p;
 
    // loop until temp<=n
    while (temp <= n)
    {
 
        // add number of numbers divisible by n
        ans += n / temp;
 
        // each time multiply temp by p
        temp = temp * p;
    }
    return ans;
}
 
// Driver function
document.write(PowerOFPINnfactorial(4, 2));
 
// This code is contributed by _saurabh_jaiswal
 
</script>

                    

Kotlin

//function to find the power of p in n! in Kotlin
fun PowerOFPINnfactorial(n: Int, p: Int)
{
    // initializing answer
    var ans = 0;
 
    // initializing
    var temp = p;
 
    // loop until temp<=n
    while(temp<=n)
    {
        // add number of numbers divisible by temp
        ans+=n/temp;
         
        // each time multiply temp by p
        temp*=p;
    }
 
    println(ans)
}
 
//Driver Code
fun main(args: Array<String>)
{
    val n = 4
    val p = 2
    PowerOFPINnfactorial(n,p)
}

                    

Output: 

3


Time Complexity :O(log_p       (n))
Auxiliary Space: O(1)


 



Last Updated : 23 Jun, 2022
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