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 <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 |
Output:
4 17 3 6 100 2