Skip to content
Related Articles

Related Articles

Improve Article

Number of distinct prime factors of first n natural numbers

  • Difficulty Level : Medium
  • Last Updated : 26 May, 2021

In this article, we study an optimized way to calculate the distinct prime factorization up to n natural number using O O(n*log n) time complexity with pre-computation allowed.
Prerequisites: Sieve of Eratosthenes, Least prime factor of numbers till n.
 

 

Key Concept: Our idea is to store the Smallest Prime Factor(SPF) for every number. Then to calculate the distinct prime factorization of the given number by dividing the given number recursively with its smallest prime factor till it becomes 1. 
 

To calculate to smallest prime factor for every number we will use the sieve of eratosthenes. In original Sieve, every time we mark a number as not-prime, we store the corresponding smallest prime factor for that number (Refer this article for better understanding).
The implementation for the above method is given below :
 

C++




// C++ program to find prime factorization upto n natural number
// O(n*Log n) time with precomputation
#include <bits/stdc++.h>
using namespace std;
#define MAXN 100001
 
// Stores smallest prime factor for every number
int spf[MAXN];
 
// Adjacency vector to store distinct prime factors
vector<int>adj[MAXN];
 
// Calculating SPF (Smallest Prime Factor) for every
// number till MAXN.
// Time Complexity : O(nloglogn)
void sieve()
{
    spf[1] = 1;
        // marking smallest prime factor for every
        // number to be itself.
    for (int i=2; i<MAXN; i++)
        spf[i] = i;
 
 
    for (int i=2; i*i<MAXN; i++)
    {
        // checking if i is prime
        if (spf[i] == i)
        {
            // marking SPF for all numbers divisible by i
            for (int j=i*i; j<MAXN; j+=i)
 
                // marking spf[j] if it is not
                // previously marked
                if (spf[j]==j)
                    spf[j] = i;
        }
    }
}
 
// A O(nlog n) function returning distinct primefactorization
// upto n natural number by dividing by smallest prime factor
// at every step
void getdistinctFactorization(int n)
{
    int index,x,i;
    for(int i=1;i<=n;i++)
    {
        index=1;
        x=i;
        if(x!=1)
            adj[i].push_back(spf[x]);
        x=x/spf[x];
        // Push all distinct prime factor in adj
        while (x != 1)
        {
            if (adj[i][index-1]!=spf[x])
            {
                adj[i].push_back(spf[x]);
                index+=1;
            }
            x = x / spf[x];
        }
    }
}
 
// Driver code
int main()
{
    // Precalculating smallest prime factor
    sieve();
     
    int n = 10;
 
     
    getdistinctFactorization(n);
     
    // Print the prime count
    cout <<"Distinct prime factor for first " << n
         <<" natural number" <<" : ";
     
    for (int i=1; i<=n; i++)
        cout << adj[i].size() << " ";
     
    return 0;
}

Java




// Java program to find prime factorization upto n natural number
// O(n*Log n) time with precomputation
import java.io.*;
import java.util.*;
class GFG
{
    static int MAXN = 100001;
     
    // Stores smallest prime factor for every number
    static int[] spf = new int[MAXN];
   
    // Adjacency vector to store distinct prime factors
    static ArrayList<ArrayList<Integer>> adj =
      new ArrayList<ArrayList<Integer>>();
     
    // Calculating SPF (Smallest Prime Factor) for every
    // number till MAXN.
    // Time Complexity : O(nloglogn)
    static void sieve()
    {
        for(int i = 0; i < MAXN; i++)
        {
            adj.add(new ArrayList<Integer>());
        }
        spf[1] = 1;
       
        // marking smallest prime factor for every
        // number to be itself.
        for (int i = 2; i < MAXN; i++)
        {
            spf[i] = i;
        }
        for (int i = 2; i * i < MAXN; i++)
        {
            // checking if i is prime
        if (spf[i] == i)
        {
                // marking SPF for all numbers divisible by i
                for (int j = i * i; j < MAXN; j += i)
                {
                   
                    // marking spf[j] if it is not
                    // previously marked
                    if (spf[j] == j)
                        spf[j] = i;
                }
            }
        }
    }
     
    // A O(nlog n) function returning distinct primefactorization
    // upto n natural number by dividing by smallest prime factor
    // at every step   
    static void getdistinctFactorization(int n)
    {
        int index, x, i;
        for(i = 1; i <= n; i++)
        {
            index = 1;
            x = i;
            if(x != 1)
                adj.get(i).add(spf[x]);
            x = x / spf[x];
           
            // Push all distinct prime factor in adj
            while (x != 1)
            {
                if (adj.get(i).get(index - 1) != spf[x])
                {
                    adj.get(i).add(spf[x]);
                    index += 1;
                }
                x = x / spf[x];
            }
        }
    }
   
    // Driver code
    public static void main (String[] args)
    {
       
        // Precalculating smallest prime factor
        sieve();    
        int n = 10;    
        getdistinctFactorization(n);
      
        // Print the prime count
        System.out.print("Distinct prime factor for first " +
                         n + " natural number" + " : ");
        for (int i = 1; i <= n; i++)
        {
            System.out.print(adj.get(i).size()+ " ");
        }
    }
}
 
// This code is contributed by avanitrachhadiya2155

Python3




# Python3 program to find prime factorization upto n natural number
# O(n*Log n) time with precomputation
 
# Calculating SPF (Smallest Prime Factor) for every
# number till MAXN.
# Time Complexity : O(nloglogn)
def sieve():
    global spf, adj, MAXN
    spf[1] = 1
     
    # marking smallest prime factor for every
    # number to be itself.
    for i in range(2, MAXN):
        spf[i] = i
 
    for i in range(2, MAXN):
        if i * i > MAXN:
            break
             
        # checking if i is prime
        if (spf[i] == i):
           
            # marking SPF for all numbers divisible by i
            for j in range(i * i, MAXN, i):
 
                # marking spf[j] if it is not
                # previously marked
                if (spf[j] == j):
                    spf[j] = i
 
# A O(nlog n) function returning distinct primefactorization
# upto n natural number by dividing by smallest prime factor
# at every step
def getdistinctFactorization(n):
    global adj, spf, MAXN
    index = 0
    for i in range(1, n + 1):
        index = 1
        x = i
        if(x != 1):
            adj[i].append(spf[x])
        x = x // spf[x]
         
        # Push all distinct prime factor in adj
        while (x != 1):
            if (adj[i][index - 1] != spf[x]):
                adj[i].append(spf[x])
                index += 1
            x = x // spf[x]
 
# Driver code
if __name__ == '__main__':
    MAXN = 100001
    spf = [0 for i in range(MAXN)]
    adj = [[] for i in range(MAXN)]
     
    # Precalculating smallest prime factor
    sieve()
    n = 10
    getdistinctFactorization(n)
 
    # Prthe prime count
    print("Distinct prime factor for first ", n, " natural number : ", end = "")
 
    for i in range(1, n + 1):
        print(len(adj[i]), end = " ")
 
# This code is contributed by mohit kumar 29

C#




using System;
using System.Collections.Generic;
public class GFG
{
  static int MAXN = 100001;
 
  // Stores smallest prime factor for every number
  static int[] spf = new int[MAXN];
 
  // Adjacency vector to store distinct prime factors
  static List<List<int>> adj = new List<List<int>>();
 
  // Calculating SPF (Smallest Prime Factor) for every
  // number till MAXN.
  // Time Complexity : O(nloglogn)
  static void sieve()
  {
    for(int i = 0; i < MAXN; i++)
    {
      adj.Add(new List<int>());
    }
    spf[1] = 1;
 
    // marking smallest prime factor for every
    // number to be itself.
    for (int i = 2; i < MAXN; i++)
    {
      spf[i] = i;
    }
    for (int i = 2; i * i < MAXN; i++)
    {
 
      // checking if i is prime
      if (spf[i] == i)
      {
        // marking SPF for all numbers divisible by i
        for (int j = i * i; j < MAXN; j += i)
        {
 
          // marking spf[j] if it is not
          // previously marked
          if (spf[j] == j)
            spf[j] = i;
        }
      }
    }
  }
 
  // A O(nlog n) function returning distinct primefactorization
  // upto n natural number by dividing by smallest prime factor
  // at every step   
  static void getdistinctFactorization(int n)
  {
    int index, x, i;
    for(i = 1; i <= n; i++)
    {
      index = 1;
      x = i;
      if(x != 1)
      {
        adj[i].Add(spf[x]);
 
      }
      x = x / spf[x];
 
      // Push all distinct prime factor in adj
      while (x != 1)
      {
        if (adj[i][index-1] != spf[x])
        {
          adj[i].Add(spf[x]);
          index += 1;
        }
        x = x / spf[x];
      }
    }
  }
 
  // Driver code
  static public void Main ()
  {
 
    // Precalculating smallest prime factor
    sieve();    
    int n = 10;    
    getdistinctFactorization(n);
 
    // Print the prime count
    Console.Write("Distinct prime factor for first " +
                  n + " natural number" + " : ");
 
    for (int i = 1; i <= n; i++)
    {
      Console.Write(adj[i].Count + " ");
    }
  }
}
 
// This code is contributed by rag2127

Javascript




<script>
// Javascript program to find prime
// factorization upto n natural number
// O(n*Log n) time with precomputation
 
    let MAXN = 100001;
    // Stores smallest prime factor
    // for every number
    let spf = new Array(MAXN);
     
    // Adjacency vector to store distinct prime factors
    let adj=[];
     
    // Calculating SPF (Smallest Prime Factor) for every
    // number till MAXN.
    // Time Complexity : O(nloglogn)
    function sieve()
    {
        for(let i = 0; i < MAXN; i++)
        {
            adj.push([]);
        }
        spf[1] = 1;
        
        // marking smallest prime factor for every
        // number to be itself.
        for (let i = 2; i < MAXN; i++)
        {
            spf[i] = i;
        }
        for (let i = 2; i * i < MAXN; i++)
        {
            // checking if i is prime
        if (spf[i] == i)
        {
                // marking SPF for all numbers divisible by i
                for (let j = i * i; j < MAXN; j += i)
                {
                    
                    // marking spf[j] if it is not
                    // previously marked
                    if (spf[j] == j)
                        spf[j] = i;
                }
            }
        }
    }
     
    // A O(nlog n) function returning
    // distinct primefactorization
    // upto n natural number by dividing by
    // smallest prime factor
    // at every step  
    function getdistinctFactorization(n)
    {
        let index, x, i;
        for(i = 1; i <= n; i++)
        {
            index = 1;
            x = i;
            if(x != 1)
                adj[i].push(spf[x]);
            x = Math.floor(x / spf[x]);
            
            // Push all distinct prime factor in adj
            while (x != 1)
            {
                if (adj[i][index - 1] != spf[x])
                {
                    adj[i].push(spf[x]);
                    index += 1;
                }
                x = Math.floor(x / spf[x]);
            }
        }
    }
     
    // Driver code
     
    // Precalculating smallest prime factor
        sieve();   
        let n = 10;   
        getdistinctFactorization(n);
       
        // Print the prime count
        document.write("Distinct prime factor for first " +
                         n + " natural number" + " : ");
        for (let i = 1; i <= n; i++)
        {
            document.write(adj[i].length+ " ");
        }
     
     
// This code is contributed by unknown2108
 
</script>
 
    
Output
Distinct prime factor for first 10 natural number : 0 1 1 1 1 2 1 1 1 2 




My Personal Notes arrow_drop_up
Recommended Articles
Page :