Given a number N which is prime. The task is to find all the numbers less than or equal to 10^6 whose minimum prime factor is N.
Examples:
Input: N = 2 Output: 500000 Input: N = 3 Output: 166667
Approach: Use sieve of Eratosthenes to find the solution to the problem. Store all the prime numbers less than 10^6 . Form another sieve that will store the count of all the numbers whose minimum prime factor is the index of the sieve. Then display the count of the prime number N (i.e. sieve_count[n]+1), where n is the prime number.
Below is the implementation of above approach:
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;#define MAX 1000000// the sieve of prime number and// count of minimum prime factorint sieve_Prime[MAX + 4] = { 0 }, sieve_count[MAX + 4] = { 0 };// form the prime sievevoid form_sieve(){ // 1 is not a prime number sieve_Prime[1] = 1; // form the sieve for (int i = 2; i <= MAX; i++) { // if i is prime if (sieve_Prime[i] == 0) { for (int j = i * 2; j <= MAX; j += i) { // if i is the least prime factor if (sieve_Prime[j] == 0) { // mark the number j as non prime sieve_Prime[j] = 1; // count the numbers whose least prime factor is i sieve_count[i]++; } } } }}// Driver codeint main(){ // form the sieve form_sieve(); int n = 2; // display cout << "Count = " << (sieve_count[n] + 1) << endl; n = 3; // display cout << "Count = " << (sieve_count[n] + 1) << endl; return 0;} |
Java
// Java implementation of above approachimport java.io.*;class GFG { static int MAX = 1000000;// the sieve of prime number and// count of minimum prime factorstatic int sieve_Prime[] = new int[MAX + 4];static int sieve_count[] = new int[MAX + 4];// form the prime sievestatic void form_sieve(){ // 1 is not a prime number sieve_Prime[1] = 1; // form the sieve for (int i = 2; i <= MAX; i++) { // if i is prime if (sieve_Prime[i] == 0) { for (int j = i * 2; j <= MAX; j += i) { // if i is the least prime factor if (sieve_Prime[j] == 0) { // mark the number j as non prime sieve_Prime[j] = 1; // count the numbers whose least prime factor is i sieve_count[i]++; } } } }}// Driver code public static void main (String[] args) { // form the sieve form_sieve(); int n = 2; // display System.out.println( "Count = " + (sieve_count[n] + 1)); n = 3; // display System.out.println ("Count = " +(sieve_count[n] + 1)); }}// This code was contributed// by inder_mca |
Python3
# Python3 implementation of# above approachMAX = 1000000# the sieve of prime number and# count of minimum prime factorsieve_Prime = [0 for i in range(MAX + 4)]sieve_count = [0 for i in range(MAX + 4)]# form the prime sievedef form_sieve(): # 1 is not a prime number sieve_Prime[1] = 1 # form the sieve for i in range(2, MAX + 1): # if i is prime if sieve_Prime[i] == 0: for j in range(i * 2, MAX + 1, i): # if i is the least prime factor if sieve_Prime[j] == 0: # mark the number j # as non prime sieve_Prime[j] = 1 # count the numbers whose # least prime factor is i sieve_count[i] += 1# Driver code# form the sieveform_sieve()n = 2# displayprint("Count =", sieve_count[n] + 1)n = 3# displayprint("Count =", sieve_count[n] + 1)# This code was contributed# by VishalBachchas |
C#
// C# implementation of above approachusing System;class GFG { static int MAX = 1000000;// the sieve of prime number and// count of minimum prime factorstatic int []sieve_Prime = new int[MAX + 4];static int []sieve_count = new int[MAX + 4];// form the prime sievestatic void form_sieve(){ // 1 is not a prime number sieve_Prime[1] = 1; // form the sieve for (int i = 2; i <= MAX; i++) { // if i is prime if (sieve_Prime[i] == 0) { for (int j = i * 2; j <= MAX; j += i) { // if i is the least prime factor if (sieve_Prime[j] == 0) { // mark the number j as non prime sieve_Prime[j] = 1; // count the numbers whose least prime factor is i sieve_count[i]++; } } } }}// Driver code public static void Main () { // form the sieve form_sieve(); int n = 2; // display Console.WriteLine( "Count = " + (sieve_count[n] + 1)); n = 3; // display Console.WriteLine ("Count = " +(sieve_count[n] + 1)); }}// This code was contributed// by shs |
PHP
<?php // PHP implementation of above approach$MAX = 1000000;// the sieve of prime number and// count of minimum prime factor$sieve_Prime = array_fill(0, $MAX + 4, NULL);$sieve_count = array_fill(0, $MAX + 4, NULL);// form the prime sievefunction form_sieve(){ global $sieve_Prime, $sieve_count, $MAX; // 1 is not a prime number $sieve_Prime[1] = 1; // form the sieve for ($i = 2; $i <= $MAX; $i++) { // if i is prime if ($sieve_Prime[$i] == 0) { for ($j = $i * 2; $j <= $MAX; $j += $i) { // if i is the least prime factor if ($sieve_Prime[$j] == 0) { // mark the number j as non prime $sieve_Prime[$j] = 1; // count the numbers whose least // prime factor is i $sieve_count[$i]++; } } } }}// Driver code// form the sieveform_sieve();$n = 2;// displayecho "Count = " . ($sieve_count[$n] + 1) . "\n";$n = 3;// displayecho "Count = " . ($sieve_count[$n] + 1) . "\n";// This code is contributed by ita_c?> |
Javascript
<script>// Javascript implementation of above approach var MAX = 1000000;// the sieve of prime number and// count of minimum prime factorvar sieve_Prime = Array.from({length: MAX + 4},(_, i) => 0);var sieve_count = Array.from({length: MAX + 4},(_, i) => 0);// form the prime sievefunction form_sieve(){ // 1 is not a prime number sieve_Prime[1] = 1; // form the sieve for (i = 2; i <= MAX; i++) { // if i is prime if (sieve_Prime[i] == 0) { for (j = i * 2; j <= MAX; j += i) { // if i is the least prime factor if (sieve_Prime[j] == 0) { // mark the number j as non prime sieve_Prime[j] = 1; // count the numbers whose least // prime factor is i sieve_count[i]++; } } } }}// Driver code// form the sieveform_sieve();var n = 2;// displaydocument.write( "Count = " + (sieve_count[n] + 1));n = 3;// displaydocument.write("<br>Count = " +(sieve_count[n] + 1)); // This code contributed by shikhasingrajput</script> |
Output
Count = 500000 Count = 166667
Time Complexity: O(N*log(log(N))), where N=106.
Auxiliary Space: O(106)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
