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Number of distinct prime factors of first n natural numbers
  • Difficulty Level : Medium
  • Last Updated : 30 Dec, 2020

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++

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// 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;
}

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Python3

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# 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

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