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Minimize swaps required to make all prime-indexed elements as prime

  • Last Updated : 08 Oct, 2021

Given an array arr[] of size N. The task is to find the minimum number of swaps required to re-arrange the array such that all prime-indexed elements are prime, If the task can’t be achieved, print “-1

Examples:

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Input:  N = 5, arr[] = {1, 2, 3, 4, 5}
Output: 0
Explanation: All the prime indices {2, 3, 5} (one-based indexing) have prime elements present on them. Therefore, minimum swaps required is 0.



Input:  N = 5, arr[] = {2, 7, 8, 5, 13}
Output: 1
Explanation: swap 8 and 5 once, to get all the prime numbers at prime indices.

 

Approach: The task can be solved using the Sieve of Eratosthenes

  • Iterate over the array and check whether the current index is prime or not, if it is a prime, check the element present on this index.
  • On observation, it can be seen that It is possible to achieve the required configuration, only if the total number of primes in the array is greater than or equal to the total number of prime indices in the array.
  • If this condition holds, then the minimum number of swaps is equal to the number of prime indices that do not have a prime number on them.

Below is the implementation of the above approach:

C++




#include <bits/stdc++.h>
using namespace std;
 
const int mxn = 1e4 + 1;
bool prime[mxn + 1];
 
// Function to pre-calculate the prime[]
// prime[i] denotes whether
// i is prime or not
void SieveOfEratosthenes()
{
    memset(prime, true, sizeof(prime));
 
    for (int p = 2; p * p <= mxn; p++) {
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true) {
            // Update all multiples
            // of p greater than or
            // equal to the square of it
            // numbers which are multiple
            // of p and are less than p^2
            // are already been marked.
            for (int i = p * p; i <= mxn; i += p)
                prime[i] = false;
        }
    }
}
 
// Function to count minimum number
// of swaps required
int countMin(int arr[], int n)
{
    // To count the minimum number of swaps
    // required to convert the array into
    // perfectly prime
    int cMinSwaps = 0;
 
    // To count total number of prime
    // indexes in the array
    int cPrimeIndices = 0;
 
    // To count the total number of
    // prime numbers in the array
 
    int cPrimeNos = 0;
 
    for (int i = 0; i < n; i++) {
        // Check whether index
        // is prime or not
 
        if (prime[i + 1]) {
            cPrimeIndices++;
 
            // Element is not prime
            if (!prime[arr[i]])
                cMinSwaps++;
            else
                cPrimeNos++;
        }
        else if (prime[arr[i]]) {
            cPrimeNos++;
        }
    }
 
    // If the total number of prime numbers
    // is greater than or equal to the total
    // number of prime indices, then it is
    // possible to convert the array into
    // perfectly prime
    if (cPrimeNos >= cPrimeIndices)
        return cMinSwaps;
    else
        return -1;
}
// Driver Code
int main()
{
    // Pre-calculate prime[]
    SieveOfEratosthenes();
 
    int n = 5;
    int arr[5] = { 2, 7, 8, 5, 13 };
 
    cout << countMin(arr, n);
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
 
class GFG {
    static boolean prime[] = new boolean[(int)1e4 + 2];
    static void SieveOfEratosthenes()
    {
        // Create a boolean array "prime[0..n]" and
        // initialize all entries it as true. A value in
        // prime[i] will finally be false if i is Not a
        // prime, else true.
 
        for (int i = 0; i < (int)1e4 + 2; i++)
            prime[i] = true;
 
        for (int p = 2; p * p < (int)1e4 + 2; p++)
        {
           
            // If prime[p] is not changed, then it is a
            // prime
            if (prime[p] == true)
            {
               
                // Update all multiples of p
                for (int i = p * p; i < (int)1e4 + 2;
                     i += p)
                    prime[i] = false;
            }
        }
    }
   
    // Function to count minimum number
    // of swaps required
    static int countMin(int[] arr, int n)
    {
       
        // To count the minimum number of swaps
        // required to convert the array into
        // perfectly prime
        int cMinSwaps = 0;
 
        // To count total number of prime
        // indexes in the array
        int cPrimeIndices = 0;
 
        // To count the total number of
        // prime numbers in the array
 
        int cPrimeNos = 0;
 
        for (int i = 0; i < n; i++) {
            // Check whether index
            // is prime or not
 
            if (prime[i + 1]) {
                cPrimeIndices++;
 
                // Element is not prime
                if (prime[arr[i]] == false)
                    cMinSwaps++;
                else
                    cPrimeNos++;
            }
            else if (prime[arr[i]]) {
                cPrimeNos++;
            }
        }
 
        // If the total number of prime numbers
        // is greater than or equal to the total
        // number of prime indices, then it is
        // possible to convert the array into
        // perfectly prime
        if (cPrimeNos >= cPrimeIndices)
            return cMinSwaps;
        else
            return -1;
    }
   
    // Driver Code
    public static void main(String[] args)
    {
       
        // Pre-calculate prime[]
        SieveOfEratosthenes();
 
        int n = 5;
        int arr[] = { 2, 7, 8, 5, 13 };
 
        System.out.println(countMin(arr, n));
    }
}
 
// This code is contributed by Potta Lokesh

Python3




# Python program for the above approach
import math
 
mxn = 10000 + 1
prime = [True for _ in range(mxn + 1)]
 
# Function to pre-calculate the prime[]
# prime[i] denotes whether
# i is prime or not
def SieveOfEratosthenes():
    global prime
    for p in range(2, int(math.sqrt(mxn)) + 1):
       
                # If prime[p] is not changed,
                # then it is a prime
        if (prime[p] == True):
                        # Update all multiples
                        # of p greater than or
                        # equal to the square of it
                        # numbers which are multiple
                        # of p and are less than p^2
                        # are already been marked.
 
            for i in range(p*p, mxn+1, p):
                prime[i] = False
 
 
# Function to count minimum number
# of swaps required
def countMin(arr, n):
 
        # To count the minimum number of swaps
        # required to convert the array into
        # perfectly prime
    cMinSwaps = 0
 
    # To count total number of prime
    # indexes in the array
    cPrimeIndices = 0
 
    # To count the total number of
    # prime numbers in the array
 
    cPrimeNos = 0
 
    for i in range(0, n):
                # Check whether index
                # is prime or not
 
        if (prime[i + 1]):
            cPrimeIndices += 1
 
            # Element is not prime
            if (not prime[arr[i]]):
                cMinSwaps += 1
            else:
                cPrimeNos += 1
 
        elif (prime[arr[i]]):
            cPrimeNos += 1
 
    # If the total number of prime numbers
    # is greater than or equal to the total
    # number of prime indices, then it is
    # possible to convert the array into
    # perfectly prime
    if (cPrimeNos >= cPrimeIndices):
        return cMinSwaps
    else:
        return -1
 
# Driver Code
if __name__ == "__main__":
 
    # Pre-calculate prime[]
    SieveOfEratosthenes()
 
    n = 5
    arr = [2, 7, 8, 5, 13]
 
    print(countMin(arr, n))
 
    # This code is contributed by rakeshsahni

C#




// C# program for the above approach
using System;
 
class GFG
{
    static bool[] prime = new bool[(int)1e4 + 2];
    static void SieveOfEratosthenes()
    {
       
        // Create a boolean array "prime[0..n]" and
        // initialize all entries it as true. A value in
        // prime[i] will finally be false if i is Not a
        // prime, else true.
        for (int i = 0; i < (int)1e4 + 2; i++)
            prime[i] = true;
 
        for (int p = 2; p * p < (int)1e4 + 2; p++) {
 
            // If prime[p] is not changed, then it is a
            // prime
            if (prime[p] == true) {
 
                // Update all multiples of p
                for (int i = p * p; i < (int)1e4 + 2;
                     i += p)
                    prime[i] = false;
            }
        }
    }
 
    // Function to count minimum number
    // of swaps required
    static int countMin(int[] arr, int n)
    {
 
        // To count the minimum number of swaps
        // required to convert the array into
        // perfectly prime
        int cMinSwaps = 0;
 
        // To count total number of prime
        // indexes in the array
        int cPrimeIndices = 0;
 
        // To count the total number of
        // prime numbers in the array
 
        int cPrimeNos = 0;
 
        for (int i = 0; i < n; i++) {
            // Check whether index
            // is prime or not
 
            if (prime[i + 1]) {
                cPrimeIndices++;
 
                // Element is not prime
                if (prime[arr[i]] == false)
                    cMinSwaps++;
                else
                    cPrimeNos++;
            }
            else if (prime[arr[i]]) {
                cPrimeNos++;
            }
        }
 
        // If the total number of prime numbers
        // is greater than or equal to the total
        // number of prime indices, then it is
        // possible to convert the array into
        // perfectly prime
        if (cPrimeNos >= cPrimeIndices)
            return cMinSwaps;
        else
            return -1;
    }
 
    // Driver Code
    public static void Main()
    {
 
        // Pre-calculate prime[]
        SieveOfEratosthenes();
 
        int n = 5;
        int[] arr = { 2, 7, 8, 5, 13 };
 
        Console.Write(countMin(arr, n));
    }
}
 
// This code is contributed by subhammahato348.

Javascript




<script>
const mxn = 1e4 + 1;
let prime = new Array(mxn + 1);
 
// Function to pre-calculate the prime[]
// prime[i] denotes whether
// i is prime or not
function SieveOfEratosthenes() {
  prime.fill(true);
 
  for (let p = 2; p * p <= mxn; p++)
  {
   
    // If prime[p] is not changed,
    // then it is a prime
    if (prime[p] == true)
    {
     
      // Update all multiples
      // of p greater than or
      // equal to the square of it
      // numbers which are multiple
      // of p and are less than p^2
      // are already been marked.
      for (let i = p * p; i <= mxn; i += p) prime[i] = false;
    }
  }
}
 
// Function to count minimum number
// of swaps required
function countMin(arr, n)
{
 
  // To count the minimum number of swaps
  // required to convert the array into
  // perfectly prime
  let cMinSwaps = 0;
 
  // To count total number of prime
  // indexes in the array
  let cPrimeIndices = 0;
 
  // To count the total number of
  // prime numbers in the array
 
  let cPrimeNos = 0;
 
  for (let i = 0; i < n; i++) {
    // Check whether index
    // is prime or not
 
    if (prime[i + 1]) {
      cPrimeIndices++;
 
      // Element is not prime
      if (!prime[arr[i]]) cMinSwaps++;
      else cPrimeNos++;
    } else if (prime[arr[i]]) {
      cPrimeNos++;
    }
  }
 
  // If the total number of prime numbers
  // is greater than or equal to the total
  // number of prime indices, then it is
  // possible to convert the array into
  // perfectly prime
  if (cPrimeNos >= cPrimeIndices) return cMinSwaps;
  else return -1;
}
 
// Driver Code
 
// Pre-calculate prime[]
SieveOfEratosthenes();
 
let n = 5;
let arr = [2, 7, 8, 5, 13];
 
document.write(countMin(arr, n));
 
// This code is contributed by gfgking.
</script>
Output
1

Time Complexity: O(mxn(log(log(mxn)))
Auxiliary Space: O(mxn)




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