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# 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++

 `// C++ implementation of the approach``#include ``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 v = ``new` `Vector();` `        ``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

 ``

Output

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