Aquarium

 Coding Problem Keys 

 Aquarium 

 Problem Statement 

An aquarium has N number of cells in it. The number o fish in each cell is represented in the form of Ai of array A, where i represents the i th index of the array A.

Shyam wants to sell the aquarium, but to sell it at a higher price, he will have to make it look better. The aquarium can be made better if in each cell there is a prime number of fish. Shyam can select any index of the array i.e. cell of the aquarium and can add a fish or can remove one fish from the aquarium. Adding or removing a fish takes him one second.

Help out Shyam so that he can make the aquarium better in the last time, and then print that time taken.

 Input Format 

• The first line contains N (Length of the array)
• The second line contains N space-separated integers.

 Output Format 

Print an integer denoting the minimum time required.

 Constraints 

1 ≤ N ≤ 10⁵

1 ≤ Ai ≤ 10000000

 Sample Input 

4

9 5 3 6

 Sample Output 

3

 Explanation 

In sample, two primes 5 and 3 are already present. You don't need to change them. For 9, you can either make 7 or 11, both takes 2 steps.

And, if you can change 6 to either 5 or 7, both takes 1 step. So, as a whole it takes 3 steps to make all elements prime.

 Note: Max Execution Time Limit: 5000 millisecs 

 Solution 

 Programming Language: Java Language 

import java.util.Scanner;
public class Main{
    static boolean isPrime(int x){
        if(x==2 || x==3 || x==5 || x==7) return true;
        else if(x%2==0 || x%3==0 || x%5==0 || x%7==0) return false;
        else if(Math.ceil(Math.sqrt(x))-Math.floor(Math.sqrt(x))==0) return false;
        else return true;
    }
    static int mintime(int A[],int N){
        int sec=0;
        int pnearl=0,pnearh=0;
        for (int i=0;i<N;i++){
            for(int j=A[i];j>=1;j--){
                if(isPrime(j)){
                    pnearl=j;
                    break;
                }
            }
             for(int j=A[i];j<=A[i]+A[i];j++){
                if(isPrime(j)){
                    pnearh=j;
                    break;
                }
            }
            if(A[i]-pnearl < pnearh-A[i]){
                sec+=A[i]-pnearl;
            }else if(A[i]-pnearl > pnearh-A[i]){
                sec+=pnearh-A[i];
            }sec+=pnearh-A[i];
        }return sec;   
    }
    public static void main(String args[]) {
      int N,A[];
      Scanner ip=new Scanner(System.in);
      N=ip.nextInt();
      A=new int[N];
      for(int i=0;i<N;i++)
        A[i]=ip.nextInt();
    System.out.println(mintime(A,N));
    }
}
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