Modify array by merging elements with addition such that it consists of only Primes.


Given an array arr[] consisting of positive integers, the task is to check whether we can modify the array by adding any of the elements of the array such that it consists of only Primes.

Examples:

Input: arr[] = {3, 5, 7, 21}
Output: YES
Explanation:
Add the following elements, 3+5+21
So the array becomes {7, 29} which consists of only primes.

Input: {2, 5, 12, 16}
Output: NO
Explanation:
There is no possible combination among elements which will make the array containing all primes

Input: {3, 5, 13, 22}
Output: YES
Explanation:
Only one combination is possible,
Add all elements – 3+5+13+22 = {43} which is only prime



Pre-requisites: Bitmasking-DP

Approach:

  • We can solve this problem using Dynamic Programming with Bitmasks. We can represent our DP state as a mask which is subset of elements.
  • So let our dp array be DP[mask] which represents whether upto this mask (the chosen elements) the subset formed will elements are primes only.
  • The max number of bits in mask will be the number of elements in the array.
  • We keep on marking the elements encoded as a sequence in mask (If i-th index element is selected, then i-th bit is set in the mask) and keep checking whether the current chosen element (current sum) is prime, if it is prime and all other elements are visited, then we return true, and we get the answer.
  • Else if other elements are not visited, then as the current sum is prime, we just need to search for other elements which can individually or by summing up to form some primes, so we can safely mark our current sum as 0 again.
  • If the mask is full (All elements are visited) and the current sum is not prime, we return false, because there is atleast one sum that is not prime.
  • Recurrence:

    DP[mask] = solve(curr + arr[i], mask | 1<<i)     
    where 0 <= i < n

Below is the implementation of the above approach.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find the 
// array of primes
#include <bits/stdc++.h>
using namespace std;
  
// DP array to store the
// ans for max 20 elements
bool dp[1 << 20];
  
// To check whether the 
// number is prime or not
bool isprime(int n)
{
    if (n == 1)
        return false;
    for (int i = 2; i * i <= n; i++)
    {
        if (n % i == 0) {
            return false;
        }
    }
    return true;
}
  
// Function to check whether the
// array can be modify so that 
// there are only primes
int solve(int arr[], int curr, 
          int mask, int n)
{
  
    // If curr is prime and all
    // elements are visited,
    // return true
    if (isprime(curr))
    {
        if (mask == (1 << n) - 1)
        {
            return true;
        }
  
        // If all elements are not 
        // visited, set curr=0, to 
        // search for new prime sum
        curr = 0;
    }
  
    // If all elements are visited
    if (mask == (1 << n) - 1)
    {
  
        // If the current sum is
        // not prime return false
        if (!isprime(curr))
        {
            return false;
        }
    }
  
    // If this state is already 
    // calculated, return the
    // answer directly
    if (dp[mask])
        return dp[mask];
  
    // Try all state of mask
    for (int i = 0; i < n; i++)
    {
        // If ith index is not set
        if (!(mask & 1 << i))
        {
            // Add the current element
            // and set ith index and recur
            if (solve(arr, curr + arr[i]
                      , mask | 1 << i, n))
            {
                // If subset can be formed
                // then return true
                return true;
            }
        }
    }
      
    // After every possibility of mask, 
    // if the subset is not formed,
    // return false by memoizing.
    return dp[mask] = false;
}
  
// Driver code
int main()
{
  
    int arr[] = { 3, 6, 7, 13 };
    int n = sizeof(arr) / sizeof(arr[0]);
      
    if(solve(arr, 0, 0, n))
    {
        cout << "YES";
    }
    else
    {
         cout << "NO";
    }
    return 0;
     
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find the array of primes
import java.util.*;
  
class GFG{
  
// dp array to store the
// ans for max 20 elements
static boolean []dp = new boolean[1 << 20];
  
// To check whether the 
// number is prime or not
static boolean isprime(int n)
{
    if (n == 1)
        return false;
  
    for(int i = 2; i * i <= n; i++)
    {
       if (n % i == 0)
       {
           return false;
       }
    }
    return true;
}
  
// Function to check whether the
// array can be modify so that 
// there are only primes
static boolean solve(int arr[], int curr, 
                     int mask, int n)
{
  
    // If curr is prime and all
    // elements are visited,
    // return true
    if (isprime(curr))
    {
        if (mask == (1 << n) - 1)
        {
            return true;
        }
  
        // If all elements are not 
        // visited, set curr=0, to 
        // search for new prime sum
        curr = 0;
    }
  
    // If all elements are visited
    if (mask == (1 << n) - 1)
    {
  
        // If the current sum is
        // not prime return false
        if (!isprime(curr))
        {
            return false;
        }
    }
  
    // If this state is already 
    // calculated, return the
    // answer directly
    if (dp[mask])
        return dp[mask];
  
    // Try all state of mask
    for(int i = 0; i < n; i++)
    {
         
       // If ith index is not set
       if ((mask & (1 << i)) == 0)
       {
             
           // Add the current element
           // and set ith index and recur
           if (solve(arr, curr + arr[i], 
                     mask | 1 << i, n))
           {
                 
               // If subset can be formed
               // then return true
               return true;
           }
       }
    }
      
    // After every possibility of mask, 
    // if the subset is not formed,
    // return false by memoizing.
    return dp[mask] = false;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 3, 6, 7, 13 };
    int n = arr.length;
      
    if(solve(arr, 0, 0, n))
    {
        System.out.print("YES");
    }
    else
    {
        System.out.print("NO");
    }
}
}
  
// This code is contributed by Rohit_ranjan

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find the array of primes
using System;
  
class GFG{
  
// dp array to store the
// ans for max 20 elements
static bool []dp = new bool[1 << 20];
  
// To check whether the 
// number is prime or not
static bool isprime(int n)
{
    if (n == 1)
        return false;
  
    for(int i = 2; i * i <= n; i++)
    {
       if (n % i == 0)
       {
           return false;
       }
    }
    return true;
}
  
// Function to check whether the
// array can be modify so that 
// there are only primes
static bool solve(int []arr, int curr, 
                  int mask, int n)
{
  
    // If curr is prime and all
    // elements are visited,
    // return true
    if (isprime(curr))
    {
        if (mask == (1 << n) - 1)
        {
            return true;
        }
  
        // If all elements are not 
        // visited, set curr=0, to 
        // search for new prime sum
        curr = 0;
    }
  
    // If all elements are visited
    if (mask == (1 << n) - 1)
    {
  
        // If the current sum is
        // not prime return false
        if (!isprime(curr))
        {
            return false;
        }
    }
  
    // If this state is already 
    // calculated, return the
    // answer directly
    if (dp[mask])
        return dp[mask];
  
    // Try all state of mask
    for(int i = 0; i < n; i++)
    {
         
       // If ith index is not set
       if ((mask & (1 << i)) == 0)
       {
             
           // Add the current element
           // and set ith index and recur
           if (solve(arr, curr + arr[i], 
                     mask | 1 << i, n))
           {
                 
               // If subset can be formed
               // then return true
               return true;
           }
       }
    }
      
    // After every possibility of mask, 
    // if the subset is not formed,
    // return false by memoizing.
    return dp[mask] = false;
}
  
// Driver code
public static void Main(String[] args)
{
    int []arr = { 3, 6, 7, 13 };
    int n = arr.Length;
      
    if(solve(arr, 0, 0, n))
    {
        Console.Write("YES");
    }
    else
    {
        Console.Write("NO");
    }
}
}
  
// This code is contributed by Rohit_ranjan

chevron_right


Output:

YES

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : Rohit_ranjan