Open In App

Sort prime numbers of an array in descending order

Improve
Improve
Like Article
Like
Save
Share
Report

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++




// 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;
}


Java




// 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


Python3




# 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


C#




// 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


Javascript




<script>
 
// Javascript implementation of the approach
 
var prime = Array(100005).fill(true);
 
function SieveOfEratosthenes( n)
{
 
    // false here indicates
    // that it is not prime
    prime[1] = false;
 
    for (var 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 (var i = p * 2; i <= n; i += p)
                prime[i] = false;
        }
    }
}
 
// Function that sorts
// all the prime numbers
// from the array in descending
function sortPrimes(arr, n)
{
    SieveOfEratosthenes(100005);
 
    // this vector will contain
    // prime numbers to sort
    var v = [];
 
    for (var i = 0; i < n; i++) {
 
        // if the element is prime
        if (prime[arr[i]])
            v.push(arr[i]);
    }
 
    v.sort((a,b)=>b-a)
 
    var j = 0;
 
    // update the array elements
    for (var i = 0; i < n; i++) {
        if (prime[arr[i]])
            arr[i] = v[j++];
    }
}
 
// Driver code
var arr = [4, 3, 2, 6, 100, 17 ];
var n = arr.length;
sortPrimes(arr, n);
// print the results.
for (var i = 0; i < n; i++) {
    document.write( arr[i] + " ");
}
 
 
</script>


Output

4 17 3 6 100 2 
  • Complexity Analysis:
  • Time Complexity: O((n * log n) + (100005)3/2)
  • Auxiliary Space: O(n + 100005)


Last Updated : 07 Sep, 2022
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads