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



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[2] = 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++

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// C++ implementation to find n'th smart number
#include<bits/stdc++.h>
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<int> result;
  
    // 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.push_back(j);
            }
        }
    }
  
    // Sort all smart numbers
    sort(result.begin(), result.end());
  
    // return n'th smart number
    return result[n-1];
}
  
// Driver program to run the case
int main()
{
    int n = 50;
    cout << smartNumber(n);
    return 0;
}

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Java

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// 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<Integer> 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

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Python3

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# 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 = [0] * 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 

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C#

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// 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<int> result = new List<int>();
  
        // 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

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PHP

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<?php
// PHP implementation to find n'th smart number 
  
// Limit on result 
$MAX = 3000; 
  
// Function to calculate n'th smart number 
function smartNumber($n
{
    global $MAX;
    // Initialize all numbers as not prime 
    $primes=array_fill(0,$MAX,0); 
  
    // iterate to mark all primes and smart number 
    $result=array(); 
  
    // Traverse all numbers till maximum limit 
    for ($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 ($j=$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) 
                    array_push($result,$j); 
            
        
    
  
    // Sort all smart numbers 
    sort($result); 
  
    // return n'th smart number 
    return $result[$n-1]; 
  
// Driver program to run the case 
  
    $n = 50; 
    echo smartNumber($n); 
   
// This code is contributed by mits 
?>

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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.

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