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Program to find the Nth Composite Number
  • Last Updated : 06 Jun, 2021

Given a positive integer N, the task is to find the Nth Composite Number.

Examples:

Input: N = 1
Output: 4

Input: N = 3 
Output: 8

 

Approach: The given problem can be solved by using the concept of Sieve of Eratosthenes. Follow the steps below to solve the problem:



  • Mark all the Prime Numbers till 105 in a boolean array say isPrime[] using Sieve of Eratosthenes.
  • Initialize an array, say composites[] that stores all the composite numbers.
  • Traverse the array isPrime[] using the variable i, if the value of isPrime[i] is false, then insert the number i in the array composites[].
  • After completing the above steps, print the value composites[N – 1] as the Nth Composite Number.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
#define MAX_SIZE 1000005
 
// Function to find the Nth Composite
// Numbers using Sieve of Eratosthenes
int NthComposite(int N)
{
    // Sieve of prime numbers
    bool IsPrime[MAX_SIZE];
 
    // Initialize the array to true
    memset(IsPrime, true, sizeof(IsPrime));
 
    // Iterate over the range [2, MAX_SIZE]
    for (int p = 2;
         p * p < MAX_SIZE; p++) {
 
        // If IsPrime[p] is true
        if (IsPrime[p] == true) {
 
            // Iterate over the
            // range [p * p, MAX_SIZE]
            for (int i = p * p;
                 i < MAX_SIZE; i += p)
                IsPrime[i] = false;
        }
    }
 
    // Stores the list of composite numbers
    vector<int> Composites;
 
    // Iterate over the range [4, MAX_SIZE]
    for (int p = 4; p < MAX_SIZE; p++)
 
        // If i is not prime
        if (!IsPrime[p])
            Composites.push_back(p);
 
    // Return Nth Composite Number
    return Composites[N - 1];
}
 
// Driver Code
int main()
{
    int N = 4;
    cout << NthComposite(N);
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
public class GFG
{
 
// Function to find the Nth Composite
// Numbers using Sieve of Eratosthenes
public static int NthComposite(int N)
{
    int MAX_SIZE = 1000005;
   
    // Sieve of prime numbers
    boolean[] IsPrime = new boolean[MAX_SIZE];
 
    // Initialize the array to true
    Arrays.fill(IsPrime, true);
 
    // Iterate over the range [2, MAX_SIZE]
    for (int p = 2; p * p < MAX_SIZE; p++) {
 
        // If IsPrime[p] is true
        if (IsPrime[p] == true) {
 
            // Iterate over the
            // range [p * p, MAX_SIZE]
            for (int i = p * p;
                 i < MAX_SIZE; i += p)
                IsPrime[i] = false;
        }
    }
 
    // Stores the list of composite numbers
    Vector<Integer> Composites = new Vector<Integer>();
 
    // Iterate over the range [4, MAX_SIZE]
    for (int p = 4; p < MAX_SIZE; p++)
 
        // If i is not prime
        if (!IsPrime[p])
            Composites.add(p);
 
    // Return Nth Composite Number
    return Composites.get(N - 1);
}
 
// Driver Code
   public static void main(String args[])
   {
     int N = 4;
     System.out.println(NthComposite(N));
    }
}
 
// This code is contributed by SoumikMondal.

Python3




# Python3 program for the above approach
 
# Function to find the Nth Composite
# Numbers using Sieve of Eratosthenes
def NthComposite(N):
   
    # Sieve of prime numbers
    IsPrime = [True]*1000005
 
    # Iterate over the range [2, 1000005]
    for p in range(2, 1000005):
        if p * p > 1000005:
            break
 
        # If IsPrime[p] is true
        if (IsPrime[p] == True):
           
            # Iterate over the
            # range [p * p, 1000005]
            for i in range(p*p,1000005,p):
                IsPrime[i] = False
 
    # Stores the list of composite numbers
    Composites = []
 
    # Iterate over the range [4, 1000005]
    for p in range(4,1000005):
 
        # If i is not prime
        if (not IsPrime[p]):
            Composites.append(p)
 
    # Return Nth Composite Number
    return Composites[N - 1]
 
# Driver Code
if __name__ == '__main__':
    N = 4
    print (NthComposite(N))
 
# This code is contributed by mohit kumar 29.

Javascript




<script>
 
// JavaScript program for the above approach
 
let MAX_SIZE =  1000005;
 
// Function to find the Nth Composite
// Numbers using Sieve of Eratosthenes
function NthComposite( N)
{
    // Sieve of prime numbers
    let IsPrime = [];
 
    // Initialize the array to true
    for(let i = 0;i<MAX_SIZE;i++)
        IsPrime.push(true);
 
    // Iterate over the range [2, MAX_SIZE]
    for (let p = 2;
         p * p < MAX_SIZE; p++) {
 
        // If IsPrime[p] is true
        if (IsPrime[p] == true) {
 
            // Iterate over the
            // range [p * p, MAX_SIZE]
            for (let i = p * p;
                 i < MAX_SIZE; i += p)
                IsPrime[i] = false;
        }
    }
 
    // Stores the list of composite numbers
    let Composites = [];
 
    // Iterate over the range [4, MAX_SIZE]
    for (let p = 4; p < MAX_SIZE; p++)
 
        // If i is not prime
        if (!IsPrime[p])
            Composites.push(p);
 
    // Return Nth Composite Number
    return Composites[N - 1];
}
 
// Driver Code
let N = 4;
document.write(NthComposite(N));
 
</script>
Output: 
9

 

Time Complexity: O(N + M * log(log(M)), where [2, M] is the range where Sieve of Eratosthenes is performed.
Auxiliary Space: O(N)

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