Sort prime numbers of an array in descending order

Given an array of integers ‘arr’, the task is to sort all the prime numbers from the array in descending order in their relative positions i.e. other positions of the other elements must not be affected.

Examples:

Input: arr[] = {2, 5, 8, 4, 3}
Output: 5 3 8 4 2

Input: arr[] = {10, 12, 2, 6, 5}
Output: 10 12 5 6 2


Approach:

  • Create a sieve to check whether an element is prime or not in O(1).
  • Traverse the array and check if the number is prime. If it is prime, store it in a vector.
  • Then, sort the vector in descending order.
  • Again traverse the array and replace the prime numbers with the vector elements one by one.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
bool prime[100005];
  
void SieveOfEratosthenes(int n)
{
  
    memset(prime, true, sizeof(prime));
  
    // false here indicates
    // that it is not prime
    prime[1] = false;
  
    for (int p = 2; p * p <= n; p++) {
  
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p]) {
  
            // Update all multiples of p,
            // set them to non-prime
            for (int i = p * 2; i <= n; i += p)
                prime[i] = false;
        }
    }
}
  
// Function that sorts
// all the prime numbers
// from the array in descending
void sortPrimes(int arr[], int n)
{
    SieveOfEratosthenes(100005);
  
    // this vector will contain
    // prime numbers to sort
    vector<int> v;
  
    for (int i = 0; i < n; i++) {
  
        // if the element is prime
        if (prime[arr[i]])
            v.push_back(arr[i]);
    }
  
    sort(v.begin(), v.end(), greater<int>());
  
    int j = 0;
  
    // update the array elements
    for (int i = 0; i < n; i++) {
        if (prime[arr[i]])
            arr[i] = v[j++];
    }
}
  
// Driver code
int main()
{
  
    int arr[] = { 4, 3, 2, 6, 100, 17 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    sortPrimes(arr, n);
  
    // print the results.
    for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
    }
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
    static boolean prime[] = new boolean[100005];
  
    static void SieveOfEratosthenes(int n)
    {
  
        Arrays.fill(prime, true);
  
        // false here indicates
        // that it is not prime
        prime[1] = false;
  
        for (int p = 2; p * p <= n; p++)
        {
  
            // If prime[p] is not changed,
            // then it is a prime
            if (prime[p]) {
  
                // Update all multiples of p,
                // set them to non-prime
                for (int i = p * 2; i < n; i += p)
                {
                    prime[i] = false;
                }
            }
        }
    }
  
    // Function that sorts
    // all the prime numbers
    // from the array in descending
    static void sortPrimes(int arr[], int n)
    {
        SieveOfEratosthenes(100005);
  
        // this vector will contain
        // prime numbers to sort
        Vector<Integer> v = new Vector<Integer>();
  
        for (int i = 0; i < n; i++)
        {
  
            // if the element is prime
            if (prime[arr[i]]) 
            {
                v.add(arr[i]);
            }
        }
        Comparator comparator = Collections.reverseOrder();
        Collections.sort(v, comparator);
  
        int j = 0;
  
        // update the array elements
        for (int i = 0; i < n; i++) 
        {
            if (prime[arr[i]]) 
            {
                arr[i] = v.get(j++);
            }
        }
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {4, 3, 2, 6, 100, 17};
        int n = arr.length;
  
        sortPrimes(arr, n);
  
        // print the results.
        for (int i = 0; i < n; i++) 
        {
            System.out.print(arr[i] + " ");
        }
    }
}
  
// This code is contributed by 29AjayKumar

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
  
def SieveOfEratosthenes(n): 
  
    # false here indicates 
    # that it is not prime 
    prime[1] = False
    p = 2
    while p * p <= n: 
  
        # If prime[p] is not changed, 
        # then it is a prime 
        if prime[p]: 
  
            # Update all multiples of p, 
            # set them to non-prime 
            for i in range(p * 2, n + 1, p): 
                prime[i] = False
          
        p += 1
  
# Function that sorts all the prime 
# numbers from the array in descending 
def sortPrimes(arr, n): 
  
    SieveOfEratosthenes(100005
  
    # This vector will contain 
    # prime numbers to sort 
    v = [] 
    for i in range(0, n): 
  
        # If the element is prime 
        if prime[arr[i]]: 
            v.append(arr[i]) 
  
    v.sort(reverse = True
    j = 0
  
    # update the array elements 
    for i in range(0, n): 
        if prime[arr[i]]: 
            arr[i] = v[j]
            j += 1
              
    return arr
      
# Driver code 
if __name__ == "__main__"
  
    arr = [4, 3, 2, 6, 100, 17
    n = len(arr) 
      
    prime = [True] * 100006
    arr = sortPrimes(arr, n) 
  
    # print the results. 
    for i in range(0, n): 
        print(arr[i], end = " "
      
# This code is contributed by Rituraj Jain

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
using System.Collections.Generic; 
  
class GFG
{
  
    static bool []prime = new bool[100005];
  
    static void SieveOfEratosthenes(int n)
    {
  
        for(int i = 0; i < 100005; i++)
            prime[i] = true;
  
        // false here indicates
        // that it is not prime
        prime[1] = false;
  
        for (int p = 2; p * p <= n; p++)
        {
  
            // If prime[p] is not changed,
            // then it is a prime
            if (prime[p]) 
            {
  
                // Update all multiples of p,
                // set them to non-prime
                for (int i = p * 2; i < n; i += p)
                {
                    prime[i] = false;
                }
            }
        }
    }
  
    // Function that sorts
    // all the prime numbers
    // from the array in descending
    static void sortPrimes(int []arr, int n)
    {
        SieveOfEratosthenes(100005);
  
        // this vector will contain
        // prime numbers to sort
        List<int> v = new List<int>();
  
        for (int i = 0; i < n; i++)
        {
  
            // if the element is prime
            if (prime[arr[i]]) 
            {
                v.Add(arr[i]);
            }
        }
        v.Sort();
        v.Reverse();
  
        int j = 0;
  
        // update the array elements
        for (int i = 0; i < n; i++) 
        {
            if (prime[arr[i]]) 
            {
                arr[i] = v[j++];
            }
        }
    }
  
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {4, 3, 2, 6, 100, 17};
        int n = arr.Length;
  
        sortPrimes(arr, n);
  
        // print the results.
        for (int i = 0; i < n; i++) 
        {
            Console.Write(arr[i] + " ");
        }
    }
}
  
// This code contributed by Rajput-Ji

chevron_right


Output:

4 17 3 6 100 2


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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 Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.