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)