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Finding powers of any number P in N!

  • Last Updated : 11 Sep, 2021

Prerequisite: Print all prime factors and their powers 
Given natural numbers N and P, the task is to find the power of P in the factorization of N!.

Examples 

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Input: N = 4, P = 2 
Output:
Explanation: 
Power of 2 in the prime factorization of 4! = 24 is 3



Input: N = 24, P = 4 
Output: 11 

Naive Approach: The idea is to find the power of P for each number from 1 to N and add them as we know during multiplication power is added. 

Time Complexity: O(N*P)

Efficient Approach: 
To find the power of the number P in N! do the following: 

  1. Find all the Prime Factors of the number P with their frequency by using the approach discussed in this article. Store the Prime Factors with their frequency in the map.
  2. Find the power of every Prime Factors of P in the factorization of N! by using the approach discussed in this article.
  3. Divide every power obtained in the above steps by their corresponding frequency in the map.
  4. Store the result of the above steps in an array, and a minimum of those elements will give the power of P in the factorization of N!.

Below is the implementation of the above approach: 

C++




// C++ program to find the power
// of P in N!
#include <bits/stdc++.h>
using namespace std;
 
// Map to store all the prime
// factors of P
unordered_map<int, int> Map;
 
// Function to find the prime
// factors of N im Map
void findPrimeFactors(int N)
{
    int i;
 
    // Clear map
    Map.clear();
 
    // Check for factors of 2
    while (N % 2 == 0) {
        Map[2] += 1;
        N /= 2;
    }
 
    // Find all the prime factors
    for (i = 3; i <= sqrt(N); i += 2) {
 
        // If i is a factors
        // then increase the
        // frequency of i
        while (N % i == 0) {
            Map[i] += 1;
            N /= i;
        }
    }
 
    if (N > 2) {
        Map[N] += 1;
    }
}
 
// Function to find the power
// of prime number P in N!
int PowInFactN(int N, int P)
{
    int ans = 0;
    int temp = P;
 
    // Loop until temp <= N
    while (temp <= N) {
 
        // Add the number of
        // numbers divisible
        // by N
        ans += N / temp;
 
        // Each time multiply
        // temp by P
        temp = temp * P;
    }
 
    // Returns ans
    return ans;
}
 
// Function that find the
// powers of any P in N!
int findPowers(int N, int P)
{
 
    // Find all prime factors
    // of number P
    findPrimeFactors(P);
 
    // To store the powers of
    // all prime factors
    vector<int> Powers;
 
    // Traverse the map
    for (auto& it : Map) {
 
        // Prime factor and
        // corres. powers
        int primeFac = it.first;
        int facPow = it.second;
 
        // Find power of prime
        // factor primeFac
        int p = PowInFactN(N,
                           primeFac);
 
        // Divide frequency by
        // facPow
        p /= facPow;
 
        // Store the power of
        // primeFac^facPow
        Powers.push_back(p);
    }
 
    // Return the minimum
    // element in Power array
    return *min_element(Powers.begin(),
                        Powers.end());
}
 
// Driver's Code
int main()
{
    int N = 24, P = 4;
 
    // Function to find power of
    // P in N!
    cout << findPowers(N, P);
    return 0;
}

Java




// Java program to find the power
// of P in N!
import java.util.*;
 
class GFG{
  
// Map to store all the prime
// factors of P
static HashMap<Integer,Integer> Map = new HashMap<Integer,Integer>();
  
// Function to find the prime
// factors of N im Map
static void findPrimeFactors(int N)
{
    int i;
  
    // Clear map
    Map.clear();
  
    // Check for factors of 2
    while (N % 2 == 0) {
        if(Map.containsKey(2))
            Map.put(2, Map.get(2) + 1);
        else
            Map.put(2, 1);
        N /= 2;
    }
  
    // Find all the prime factors
    for (i = 3; i <= Math.sqrt(N); i += 2) {
  
        // If i is a factors
        // then increase the
        // frequency of i
        while (N % i == 0) {
            if(Map.containsKey(i))
                Map.put(i, Map.get(i) + 1);
            else
                Map.put(i, 1);
            N /= i;
        }
    }
  
    if (N > 2) {
        if(Map.containsKey(N))
            Map.put(N, Map.get(N) + 1);
        else
            Map.put(N, 1);
    }
}
  
// Function to find the power
// of prime number P in N!
static int PowInFactN(int N, int P)
{
    int ans = 0;
    int temp = P;
  
    // Loop until temp <= N
    while (temp <= N) {
  
        // Add the number of
        // numbers divisible
        // by N
        ans += N / temp;
  
        // Each time multiply
        // temp by P
        temp = temp * P;
    }
  
    // Returns ans
    return ans;
}
  
// Function that find the
// powers of any P in N!
static int findPowers(int N, int P)
{
  
    // Find all prime factors
    // of number P
    findPrimeFactors(P);
  
    // To store the powers of
    // all prime factors
    Vector<Integer> Powers = new Vector<Integer>();
  
    // Traverse the map
    for (Map.Entry<Integer, Integer> it : Map.entrySet()) {
  
        // Prime factor and
        // corres. powers
        int primeFac = it.getKey();
        int facPow = it.getValue();
  
        // Find power of prime
        // factor primeFac
        int p = PowInFactN(N,
                           primeFac);
  
        // Divide frequency by
        // facPow
        p /= facPow;
  
        // Store the power of
        // primeFac^facPow
        Powers.add(p);
    }
  
    // Return the minimum
    // element in Power array
    return Collections.min(Powers);
}
  
// Driver's Code
public static void main(String[] args)
{
    int N = 24, P = 4;
  
    // Function to find power of
    // P in N!
    System.out.print(findPowers(N, P));
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 program to find the power
# of P in N!
import math
 
# Map to store all
# the prime factors of P
Map = {}
 
# Function to find the prime
# factors of N im Map
def findPrimeFactors(N):
 
    # Clear map
    Map.clear()
 
    # Check for factors of 2
    while (N % 2 == 0):
        if 2 in Map:
            Map[2] += 1
        else:
            Map[2] = 1
        N = N // 2
 
    # Find all the prime factors
    for i in range(3, int(math.sqrt(N)) + 1, 2):
       
        # If i is a factors
        # then increase the
        # frequency of i
        while (N % i == 0):
            if i in Map:
                Map[i] += 1
            else:
                Map[i] = 1
                 
            N = N // i
 
    if (N > 2):
        if N in Map:
            Map[N] += 1
        else:
            Map[N] = 1
 
# Function to find the power
# of prime number P in N!
def PowInFactN(N, P):
 
    ans = 0
    temp = P
 
    # Loop until temp <= N
    while (temp <= N):
 
        # Add the number of
        # numbers divisible
        # by N
        ans = ans + (N // temp)
 
        # Each time multiply
        # temp by P
        temp = temp * P
 
    # Returns ans
    return ans
 
# Function that find the
# powers of any P in N!
def findPowers(N, P):
 
    # Find all prime factors
    # of number P
    findPrimeFactors(P)
 
    # To store the powers of
    # all prime factors
    Powers = []
 
    # Traverse the map
    for it1, it2 in Map.items():
 
        # Prime factor and
        # corres. powers
        primeFac = it1
        facPow = it2
 
        # Find power of prime
        # factor primeFac
        p = PowInFactN(N, primeFac)
 
        # Divide frequency by
        # facPow
        p = p // facPow
 
        # Store the power of
        # primeFac^facPow
        Powers.append(p)
 
    # Return the minimum
    # element in Power array
    return min(Powers)
 
N, P = 24, 4
 
# Driver code
if __name__ == "__main__":
   
  # Function to find
  # power of P in N!
  print(findPowers(N, P))
 
# This code is contributed by divyeshrabadiya07

C#




// C# program to find the power
// of P in N!
using System;
using System.Linq;
using System.Collections.Generic;
 
class GFG{
 
// Map to store all the prime
// factors of P
static Dictionary<int,int> Map = new Dictionary<int,int>();
 
// Function to find the prime
// factors of N im Map
static void findPrimeFactors(int N)
{
    int i;
 
    // Clear map
    Map.Clear();
 
    // Check for factors of 2
    while (N % 2 == 0) {
        if(Map.ContainsKey(2))
            Map[2] = Map[2] + 1;
        else
            Map.Add(2, 1);
        N /= 2;
    }
 
    // Find all the prime factors
    for (i = 3; i <= Math.Sqrt(N); i += 2) {
 
        // If i is a factors
        // then increase the
        // frequency of i
        while (N % i == 0) {
            if(Map.ContainsKey(i))
                Map[i] = Map[i] + 1;
            else
                Map.Add(i, 1);
            N /= i;
        }
    }
 
    if (N > 2) {
        if(Map.ContainsKey(N))
            Map[N] =Map[N] + 1;
        else
            Map.Add(N, 1);
    }
}
 
// Function to find the power
// of prime number P in N!
static int PowInFactN(int N, int P)
{
    int ans = 0;
    int temp = P;
 
    // Loop until temp <= N
    while (temp <= N) {
 
        // Add the number of
        // numbers divisible
        // by N
        ans += N / temp;
 
        // Each time multiply
        // temp by P
        temp = temp * P;
    }
 
    // Returns ans
    return ans;
}
 
// Function that find the
// powers of any P in N!
static int findPowers(int N, int P)
{
 
    // Find all prime factors
    // of number P
    findPrimeFactors(P);
 
    // To store the powers of
    // all prime factors
    List<int> Powers = new List<int>();
 
    // Traverse the map
    foreach (KeyValuePair<int, int> it in Map) {
 
        // Prime factor and
        // corres. powers
        int primeFac = it.Key;
        int facPow = it.Value;
 
        // Find power of prime
        // factor primeFac
        int p = PowInFactN(N,
                        primeFac);
 
        // Divide frequency by
        // facPow
        p /= facPow;
 
        // Store the power of
        // primeFac^facPow
        Powers.Add(p);
    }
 
    // Return the minimum
    // element in Power array
    return Powers.Min();
}
 
// Driver's Code
public static void Main(String[] args)
{
    int N = 24, P = 4;
 
    // Function to find power of
    // P in N!
    Console.Write(findPowers(N, P));
}
}
 
// This code is contributed by sapnasingh4991
Output: 
11

 

Time Complexity: O(sqrt(P)*(logP N))




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