# 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.

**Algorithm**

Below is the implementation of the algorithm, where least_prime[] saves the value of the least prime factor corresponding to the respective index.

## C/C++

`// 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 = 2*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; ` `} ` |

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## Java

`// 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 = ` `2` `*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)> ` |

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## Python 3

`# 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` `(` `2` `*` `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. ` |

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## C#

`// 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 = 2*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 ` |

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## PHP

`<?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` `= 2 * ` `$i` `; ` ` ` `$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 ` `?> ` |

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**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(nlog(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?

This article is contributed by **Ayush Khanduri**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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