Least prime factor of numbers till n
Given a number n, print least prime factors of all numbers from 1 to n. The least prime factor of an integer n is the smallest prime number that divides the number. The least prime factor of all even numbers is 2. A prime number is its own least prime factor (as well as its own greatest prime factor).
Note: We need to print 1 for 1.
Example :
Input : 6
Output : Least Prime factor of 1: 1
Least Prime factor of 2: 2
Least Prime factor of 3: 3
Least Prime factor of 4: 2
Least Prime factor of 5: 5
Least Prime factor of 6: 2
We can use a variation of sieve of Eratosthenes to solve the above problem.
- Create a list of consecutive integers from 2 through n: (2, 3, 4, …, n).
- Initially, let i equal 2, the smallest prime number.
- Enumerate the multiples of i by counting to n from 2i in increments of i, and mark them as having least prime factor as i (if not already marked). Also mark i as least prime factor of i (i itself is a prime number).
- Find the first number greater than i in the list that is not marked. If there was no such number, stop. Otherwise, let i now equal this new number (which is the next prime), and repeat from step 3.
Below is the implementation of the algorithm, where least_prime[] saves the value of the least prime factor corresponding to the respective index.
C++
#include<bits/stdc++.h>
using namespace std;
void leastPrimeFactor( int n)
{
vector< int > least_prime(n+1, 0);
least_prime[1] = 1;
for ( int i = 2; i <= n; i++)
{
if (least_prime[i] == 0)
{
least_prime[i] = i;
for ( int j = i*i; j <= n; j += i)
if (least_prime[j] == 0)
least_prime[j] = i;
}
}
for ( int i = 1; i <= n; i++)
cout << "Least Prime factor of "
<< i << ": " << least_prime[i] << "\n" ;
}
int main()
{
int n = 10;
leastPrimeFactor(n);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG
{
public static void leastPrimeFactor( int n)
{
int [] least_prime = new int [n+ 1 ];
least_prime[ 1 ] = 1 ;
for ( int i = 2 ; i <= n; i++)
{
if (least_prime[i] == 0 )
{
least_prime[i] = i;
for ( int j = i*i; j <= n; j += i)
if (least_prime[j] == 0 )
least_prime[j] = i;
}
}
for ( int i = 1 ; i <= n; i++)
System.out.println( "Least Prime factor of " +
+ i + ": " + least_prime[i]);
}
public static void main (String[] args)
{
int n = 10 ;
leastPrimeFactor(n);
}
}
|
Python 3
def leastPrimeFactor(n) :
least_prime = [ 0 ] * (n + 1 )
least_prime[ 1 ] = 1
for i in range ( 2 , n + 1 ) :
if (least_prime[i] = = 0 ) :
least_prime[i] = i
for j in range (i * i, n + 1 , i) :
if (least_prime[j] = = 0 ) :
least_prime[j] = i
for i in range ( 1 , n + 1 ) :
print ( "Least Prime factor of "
,i , ": " , least_prime[i] )
n = 10
leastPrimeFactor(n)
|
C#
using System;
class GFG
{
public static void leastPrimeFactor( int n)
{
int []least_prime = new int [n+1];
least_prime[1] = 1;
for ( int i = 2; i <= n; i++)
{
if (least_prime[i] == 0)
{
least_prime[i] = i;
for ( int j = i*i; j <= n; j += i)
if (least_prime[j] == 0)
least_prime[j] = i;
}
}
for ( int i = 1; i <= n; i++)
Console.WriteLine( "Least Prime factor of " +
i + ": " + least_prime[i]);
}
public static void Main ()
{
int n = 10;
leastPrimeFactor(n);
}
}
|
PHP
<?php
function leastPrimeFactor( $n )
{
$least_prime = array ( $n + 1);
for ( $i = 0;
$i <= $n ; $i ++)
$least_prime [ $i ] = 0;
$least_prime [1] = 1;
for ( $i = 2; $i <= $n ; $i ++)
{
if ( $least_prime [ $i ] == 0)
{
$least_prime [ $i ] = $i ;
for ( $j = $i * $i ;
$j <= $n ; $j += $i )
if ( $least_prime [ $j ] == 0)
$least_prime [ $j ] = $i ;
}
}
for ( $i = 1; $i <= $n ; $i ++)
echo "Least Prime factor of " .
$i . ": " .
$least_prime [ $i ] . "\n" ;
}
$n = 10;
leastPrimeFactor( $n );
?>
|
Javascript
<script>
function leastPrimeFactor( n)
{
let least_prime = Array(n+1).fill(0);
least_prime[1] = 1;
for (let i = 2; i <= n; i++)
{
if (least_prime[i] == 0)
{
least_prime[i] = i;
for (let j = i*i; j <= n; j += i)
if (least_prime[j] == 0)
least_prime[j] = i;
}
}
for (let i = 1; i <= n; i++)
document.write( "Least Prime factor of "
+ i + ": " + least_prime[i] + "<br/>" );
}
let n = 10;
leastPrimeFactor(n);
</script>
|
Output
Least Prime factor of 1: 1
Least Prime factor of 2: 2
Least Prime factor of 3: 3
Least Prime factor of 4: 2
Least Prime factor of 5: 5
Least Prime factor of 6: 2
Least Prime factor of 7: 7
Least Prime factor of 8: 2
Least Prime factor of 9: 3
Least Prime factor of 10: 2
Time Complexity: O(n*log(n))
Auxiliary Space: O(n)
References:
1. https://www.geeksforgeeks.org/sieve-of-eratosthenes/
2. https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
3. https://oeis.org/wiki/Least_prime_factor_of_n
Exercise:
Can we extend this algorithm or use least_prime[] to find all the prime factors for numbers till n?
Last Updated :
25 Apr, 2023
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