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++ program to print the least prime factors // of numbers less than or equal to // n using modified Sieve of Eratosthenes #include<bits/stdc++.h> using namespace std;
void leastPrimeFactor( int n)
{ // Create a vector to store least primes.
// Initialize all entries as 0.
vector< int > least_prime(n+1, 0);
// We need to print 1 for 1.
least_prime[1] = 1;
for ( int i = 2; i <= n; i++)
{
// least_prime[i] == 0
// means it i is prime
if (least_prime[i] == 0)
{
// marking the prime number
// as its own lpf
least_prime[i] = i;
// mark it as a divisor for all its
// multiples if not already marked
for ( int j = i*i; j <= n; j += i)
if (least_prime[j] == 0)
least_prime[j] = i;
}
}
// print least prime factor of
// of numbers till n
for ( int i = 1; i <= n; i++)
cout << "Least Prime factor of "
<< i << ": " << least_prime[i] << "\n" ;
} // Driver program to test above function int main()
{ int n = 10;
leastPrimeFactor(n);
return 0;
} |
// Java program to print the least prime factors // of numbers less than or equal to // n using modified Sieve of Eratosthenes import java.io.*;
import java.util.*;
class GFG
{ public static void leastPrimeFactor( int n)
{
// Create a vector to store least primes.
// Initialize all entries as 0.
int [] least_prime = new int [n+ 1 ];
// We need to print 1 for 1.
least_prime[ 1 ] = 1 ;
for ( int i = 2 ; i <= n; i++)
{
// least_prime[i] == 0
// means it i is prime
if (least_prime[i] == 0 )
{
// marking the prime number
// as its own lpf
least_prime[i] = i;
// mark it as a divisor for all its
// multiples if not already marked
for ( int j = i*i; j <= n; j += i)
if (least_prime[j] == 0 )
least_prime[j] = i;
}
}
// print least prime factor of
// of numbers till n
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);
}
} // Code Contributed by Mohit Gupta_OMG <(0_o)> |
# Python 3 program to print the # least prime factors of numbers # less than or equal to n using # modified Sieve of Eratosthenes def leastPrimeFactor(n) :
# Create a vector to store least primes.
# Initialize all entries as 0.
least_prime = [ 0 ] * (n + 1 )
# We need to print 1 for 1.
least_prime[ 1 ] = 1
for i in range ( 2 , n + 1 ) :
# least_prime[i] == 0
# means it i is prime
if (least_prime[i] = = 0 ) :
# marking the prime number
# as its own lpf
least_prime[i] = i
# mark it as a divisor for all its
# multiples if not already marked
for j in range (i * i, n + 1 , i) :
if (least_prime[j] = = 0 ) :
least_prime[j] = i
# print least prime factor
# of numbers till n
for i in range ( 1 , n + 1 ) :
print ( "Least Prime factor of "
,i , ": " , least_prime[i] )
# Driver program n = 10
leastPrimeFactor(n) # This code is contributed # by Nikita Tiwari. |
// C# program to print the least prime factors // of numbers less than or equal to // n using modified Sieve of Eratosthenes using System;
class GFG
{ public static void leastPrimeFactor( int n)
{
// Create a vector to store least primes.
// Initialize all entries as 0.
int []least_prime = new int [n+1];
// We need to print 1 for 1.
least_prime[1] = 1;
for ( int i = 2; i <= n; i++)
{
// least_prime[i] == 0
// means it i is prime
if (least_prime[i] == 0)
{
// marking the prime number
// as its own lpf
least_prime[i] = i;
// mark it as a divisor for all its
// multiples if not already marked
for ( int j = i*i; j <= n; j += i)
if (least_prime[j] == 0)
least_prime[j] = i;
}
}
// print least prime factor of
// of numbers till n
for ( int i = 1; i <= n; i++)
Console.WriteLine( "Least Prime factor of " +
i + ": " + least_prime[i]);
}
// Driver code
public static void Main ()
{
int n = 10;
// Function calling
leastPrimeFactor(n);
}
} // This code is contributed by Nitin Mittal |
<?php // PHP program to print the // least prime factors of // numbers less than or equal // to n using modified Sieve // of Eratosthenes function leastPrimeFactor( $n )
{ // Create a vector to
// store least primes.
// Initialize all entries
// as 0.
$least_prime = array ( $n + 1);
for ( $i = 0;
$i <= $n ; $i ++)
$least_prime [ $i ] = 0;
// We need to
// print 1 for 1.
$least_prime [1] = 1;
for ( $i = 2; $i <= $n ; $i ++)
{
// least_prime[i] == 0
// means it i is prime
if ( $least_prime [ $i ] == 0)
{
// marking the prime
// number as its own lpf
$least_prime [ $i ] = $i ;
// mark it as a divisor
// for all its multiples
// if not already marked
for ( $j = $i * $i ;
$j <= $n ; $j += $i )
if ( $least_prime [ $j ] == 0)
$least_prime [ $j ] = $i ;
}
}
// print least prime
// factor of numbers
// till n
for ( $i = 1; $i <= $n ; $i ++)
echo "Least Prime factor of " .
$i . ": " .
$least_prime [ $i ] . "\n" ;
} // Driver Code $n = 10;
leastPrimeFactor( $n );
// This code is contributed // by Sam007 ?> |
<script> // javascript program to print the least prime factors // of numbers less than or equal to // n using modified Sieve of Eratosthenes function leastPrimeFactor( n)
{ // Create a vector to store least primes.
// Initialize all entries as 0.
let least_prime = Array(n+1).fill(0);
// We need to print 1 for 1.
least_prime[1] = 1;
for (let i = 2; i <= n; i++)
{
// least_prime[i] == 0
// means it i is prime
if (least_prime[i] == 0)
{
// marking the prime number
// as its own lpf
least_prime[i] = i;
// mark it as a divisor for all its
// multiples if not already marked
for (let j = i*i; j <= n; j += i)
if (least_prime[j] == 0)
least_prime[j] = i;
}
}
// print least prime factor of
// of numbers till n
for (let i = 1; i <= n; i++)
document.write( "Least Prime factor of "
+ i + ": " + least_prime[i] + "<br/>" );
} // Driver program to test above function let n = 10;
leastPrimeFactor(n);
// This code is contributed by Rajput-Ji </script> |
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/amp/
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?