Number of distinct prime factors of first n natural numbers (original) (raw)
Last Updated : 11 Jul, 2025
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 vectoradj[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(nLog 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] = iA 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)
# Print 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> adj = new List<List>();
// 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()); } 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
`
Output
Distinct prime factor for first 10 natural number : 0 1 1 1 1 2 1 1 1 2