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Power of a prime number ‘r’ in n!

Given an integer n, find the power of a given prime number(r) in n!
Examples : 
 

Input  : n = 6  r = 3
         Factorial of 6 is 720 = 2^4 * 3^2 *5 (prime factor 2,3,5)
         and power of 3 is 2 
Output : 2

Input  : n = 2000 r = 7
Output : 330

A simple method is to first calculate factorial of n, then factorize it to find the power of a prime number. 
The above method can cause overflow for a slightly bigger numbers as factorial of a number is a big number. The idea is to consider prime factors of a factorial n.
Legendre Factorization of n! 
For any prime number p and any integer n, let Vp(n) be the exponent of the largest power of p that divides n (that is, the p-adic valuation of n). Then 
Vp(n) = summation of floor(n / p^i) and i goes from 1 to infinity 
While the formula on the right side is an infinite sum, for any particular values of n and p it has only finitely many nonzero terms: for every i large enough that p^i > n, one has floor(n/p^i) = 0 . since, the sum is convergent.
 

Power of ‘r’ in n! = floor(n/r) + floor(n/r^2) + floor(n/r^3) + ....

Program to count power of a no. r in n! based on above formula. 
 




// C++ program to find power of
// a prime number ‘r’ in n!
#include <bits/stdc++.h>
using  namespace std;
 
// Function to return power of a
// no. 'r' in factorial of n
int power(int n, int r)
{          
    // Keep dividing n by powers of
    // 'r' and update count
    int count = 0;
    for (int i = r; (n / i) >= 1; i = i * r)   
        count += n / i;
    return count;
}
 
// Driver program to
// test above function
int main()
{
    int n = 6,  r = 3;  
    printf(" %d ", power(n, r));   
    return 0;
}




// Java program to find power of
// a prime number 'r' in n!
 
class GFG {
     
// Function to return power of a
// no. 'r' in factorial of n
static int power(int n, int r) {
     
    // Keep dividing n by powers of
    // 'r' and update count
    int count = 0;
    for (int i = r; (n / i) >= 1; i = i * r)
    count += n / i;
    return count;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 6, r = 3;
    System.out.print(power(n, r));
}
}
 
// This code is contributed by Anant Agarwal.




# Python3 program to find power
# of a prime number ‘r’ in n!
 
# Function to return power of a
# no. 'r' in factorial of n
def power(n, r):
         
    # Keep dividing n by powers of
    # 'r' and update count
    count = 0; i = r
     
    while((n / i) >= 1):
        count += n / i
        i = i * r
         
    return int(count)
 
# Driver Code
n = 6; r = 3
print(power(n, r))
 
# This code is contributed by Smitha Dinesh Semwal.




// C# program to find power of
// a prime number 'r' in n!
using System;
 
class GFG {
     
// Function to return power of a
// no. 'r' in factorial of n
static int power(int n, int r) {
     
    // Keep dividing n by powers of
    // 'r' and update count
    int count = 0;
    for (int i = r; (n / i) >= 1; i = i * r)
    count += n / i;
    return count;
}
 
// Driver code
public static void Main()
{
    int n = 6, r = 3;
    Console.WriteLine(power(n, r));
}
}
 
// This code is contributed by vt_m.




<?php
//PHP program to find power of
// a prime number ‘r’ in n!
 
// Function to return power of a
// no. 'r' in factorial of n
function power($n, $r)
{        
     
    // Keep dividing n by powers 
    // of 'r' and update count
    $count = 0;
    for ($i = $r; ($n / $i) >= 1;
                   $i = $i * $r)
        $count += $n / $i;
    return $count;
}
 
    // Driver Code
    $n = 6; $r = 3;
    echo power($n, $r);
 
// This code is contributed by ajit
?>




<script>
 
// JavaScript program to find power of
// a prime number 'r' in n!
 
// Function to return power of a
// no. 'r' in factorial of n
function power(n, r) {
       
    // Keep dividing n by powers of
    // 'r' and update count
    let count = 0;
    for (let i = r; (n / i) >= 1; i = i * r)
    count += n / i;
    return count;
}
 
// Driver code
 
    let n = 6, r = 3;
    document.write(power(n, r));
 
// This code is contributed by souravghosh0416.
</script>

Output: 
 

2

Time Complexity: O(log r) 
Auxiliary Space: O(1)

Please suggest if someone has a better solution which is more efficient in terms of space and time.


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