N’th Smart Number

Given a number n, find n’th smart number (1<=n<=1000). Smart number is a number which has at least three distinct prime factors. We are given an upper limit on value of result as MAX

For example 30 is 1st smart number because it has 2, 3, 5 as it’s distinct prime factors. 42 is 2nd smart number because it has 2, 3, 7 as it’s distinct prime factors.
Examples:

Input : n = 1
Output: 30
// three distinct prime factors 2, 3, 5

Input : n = 50
Output: 273
// three distinct prime factors 3, 7, 13

Input : n = 1000
Output: 2664
// three distinct prime factors 2, 3, 37

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

The idea is based on Sieve of Eratosthenes. We use an array to use an array prime[] to keep track of prime numbers. We also use the same array to keep track of the count of prime factors seen so far. Whenever the count reaches 3, we add the number to result.

• Take an array primes[] and initialize it with 0.
• Now we know that first prime number is i = 2 so mark primes = 1 i.e; primes[i] = 1 indicates that ‘i’ is prime number.
• Now traverse the primes[] array and mark all multiples of ‘i’ by condition primes[j] -= 1 where ‘j’ is multiple of ‘i’, and check the condition primes[j]+3 = 0 because whenever primes[j] become -3 it indicates that previously it had been multiple of three distinct prime factors. If condition primes[j]+3=0 becomes true that means ‘j’ is a Smart Number so store it in a array result[].
• Now sort array result[] and return result[n-1].

Below is the implementation of above idea.

C++

 // C++ implementation to find n'th smart number #include using namespace std;    // Limit on result const int MAX = 3000;    // Function to calculate n'th smart number int smartNumber(int n) {     // Initialize all numbers as not prime     int primes[MAX] = {0};        // iterate to mark all primes and smart number     vector result;        // Traverse all numbers till maximum limit     for (int i=2; i

Java

 // Java implementation to find n'th smart number import java.util.*; import java.lang.*;    class GFG {        // Limit on result     static int MAX = 3000;        // Function to calculate n'th smart number     public static int smartNumber(int n)     {                    // Initialize all numbers as not prime         Integer[] primes = new Integer[MAX];         Arrays.fill(primes, new Integer(0));            // iterate to mark all primes and smart         // number         Vector result = new Vector<>();            // Traverse all numbers till maximum         // limit         for (int i = 2; i < MAX; i++)         {                            // 'i' is maked as prime number             // because it is not multiple of             // any other prime             if (primes[i] == 0)             {                 primes[i] = 1;                    // mark all multiples of 'i'                  // as non prime                 for (int j = i*2; j < MAX; j = j+i)                 {                     primes[j] -= 1;                            // If i is the third prime                     // factor of j then add it                     // to result as it has at                     // least three prime factors.                     if ( (primes[j] + 3) == 0)                         result.add(j);                 }             }         }            // Sort all smart numbers         Collections.sort(result);            // return n'th smart number         return result.get(n-1);        }        // Driver program to run the case     public static void main(String[] args)     {         int n = 50;         System.out.println(smartNumber(n));        } }    // This code is contributed by Prasad Kshirsagar

Python3

 # Python3 implementation to find # n'th smart number     # Limit on result  MAX = 3000;     # Function to calculate n'th # smart number  def smartNumber(n):         # Initialize all numbers as not prime      primes =  * MAX;         # iterate to mark all primes      # and smart number      result = [];         # Traverse all numbers till maximum limit      for i in range(2, MAX):                     # 'i' is maked as prime number because          # it is not multiple of any other prime          if (primes[i] == 0):              primes[i] = 1;                 # mark all multiples of 'i' as non prime             j = i * 2;             while (j < MAX):                  primes[j] -= 1;                     # If i is the third prime factor of j                  # then add it to result as it has at                  # least three prime factors.                  if ( (primes[j] + 3) == 0):                      result.append(j);                 j = j + i;        # Sort all smart numbers      result.sort();         # return n'th smart number      return result[n - 1];     # Driver Code n = 50;  print(smartNumber(n));     # This code is contributed by mits

C#

 // C# implementation to find n'th smart number using System.Collections.Generic;    class GFG {        // Limit on result     static int MAX = 3000;        // Function to calculate n'th smart number     public static int smartNumber(int n)     {                    // Initialize all numbers as not prime         int[] primes = new int[MAX];            // iterate to mark all primes and smart         // number         List result = new List();            // Traverse all numbers till maximum         // limit         for (int i = 2; i < MAX; i++)         {                            // 'i' is maked as prime number             // because it is not multiple of             // any other prime             if (primes[i] == 0)             {                 primes[i] = 1;                    // mark all multiples of 'i'                  // as non prime                 for (int j = i*2; j < MAX; j = j+i)                 {                     primes[j] -= 1;                            // If i is the third prime                     // factor of j then add it                     // to result as it has at                     // least three prime factors.                     if ( (primes[j] + 3) == 0)                         result.Add(j);                 }             }         }            // Sort all smart numbers         result.Sort();            // return n'th smart number         return result[n-1];        }        // Driver program to run the case     public static void Main()     {         int n = 50;         System.Console.WriteLine(smartNumber(n));        } }    // This code is contributed by mits

PHP



Output:

273

This article is contributed by Shashak Mishra ( Gullu ). 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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes arrow_drop_up

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.