Largest subset with sum of every pair as prime

Given an array A[], find a subset of maximum size in which sum of every pair of elements is a prime number. Print its length and the subset. Consider many queries for different arrays and maximum value of an element as 100000.

Examples :

Input : A[] = {2, 1, 2}
Output : 2
1 2
Explanation :
Here, we can only form subsets with size 1 and 2.
maximum sized subset = {1, 2}, 1 + 2 = 3, which
is prime number.
So, Answer = 2 (size), {1, 2} (subset)

Input : A[] = {2, 1, 1}
Output : 3
1 1 2
Explanation :
Maximum subset = {2, 1, 2}, since 1 + 2 = 3,
1 + 1 = 2, both are prime numbers.
Answer = 3 (size), {2, 1, 1} (subset).

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Let’s make some observations and then move to problem. Sum of two numbers is even if and only both the numbers are either odd or even. An even number cannot be a prime number except 2. Now, if we take three numbers a, b and c, two of them should be either odd or even(Pigeonhole theorem). So, our solution exists only in two cases – (Let the subset be B)

• Case I : When B contains only two integers(>1) whose sum is a prime number.
• Case II : When B contains some number of ones(1s) and another number X, where X + 1 should be a prime(Only possible when X is an even number).

First count the number of ones in the array using a for loop.

1. If the count of 1s is greater than 0, then traverse the whole the array and check if [A[i] + 1] is a prime number and (A[i] != 1), if found any, print the size of subarray as (count of 1s) +1 and all the ones(1s) and the found A[i]. Exit the program.
2. If the above step fails (i.e, A[i] not found), print all the ones(1s). Exit the program.
3. If above step fails (i.e, count of 1s = 0), Check every pair of elements in the array for their sum to be a prime. Print 2 and the pair of integers.
4. Else Print -1.

Below is implementation of above approach :

C++

 // CPP program to find a subset in which sum of  // every pair in it is a prime #include using namespace std;    #define MAX 100001    bool isPrime[MAX] = { 0 };    int sieve() {     for (int p = 2; p * p < MAX; p++)      {         // If isPrime[p] is not changed, then it         // is a prime         if (isPrime[p] == 0)          {             // Update all multiples of p             for (int i = p * 2; i < MAX; i += p)                 isPrime[i] = 1;         }     } }    int findSubset(int a[], int n) {     int cnt1 = 0;        // Counting no.of ones in the array     for (int i = 0; i < n; i++)          if (a[i] == 1)             cnt1++;        // Case-I: count of ones(1s) > 0 and      // an integer > 1 is present in the array     if (cnt1 > 0)      {         for (int i = 0; i < n; i++)          {             // Find a[i], where a[i] + 1 is prime.             if ((a[i] != 1) and (isPrime[a[i] + 1] == 0))              {                 cout << cnt1 + 1 << endl;                    // Print all the ones(1s).                 for (int j = 0; j < cnt1; j++)                             cout << 1 << " ";                 cout << a[i] << endl; // print a[i].                 return 0;             }         }     }        // Case-II: array contains only ones(1s)     if (cnt1 >= 2)      {         cout << cnt1 << endl;            // Print all ones(1s).         for (int i = 0; i < cnt1; i++)              cout << 1 << " ";            cout << endl;         return 0;     }        // Case-III: array does not contain 1s     for (int i = 0; i < n; i++)      {         for (int j = i + 1; j < n; j++)          {             // Find a pair of integers whose sum is prime             if (isPrime[a[i] + a[j]] == 0)              {                 cout << 2 << endl;                 cout << a[i] << " " << a[j] << endl;                 return 0;             }         }     }        // Array contains only a single element.     cout << -1 << endl; }    // Driver function int main() {     sieve();     int A[] = { 2, 1, 1 };     int n = sizeof(A) / sizeof(A);     findSubset(A, n);     return 0; }

Java

 // Java program to find a  // subset in which sum of  // every pair in it is a prime import java.io.*;    class GFG {     static int MAX = 100001;            static int []isPrime = new int[MAX];            static int sieve()     {         for (int p = 2;                   p * p < MAX; p++)          {             // If isPrime[p] is              // not changed, then             // it is a prime             if (isPrime[p] == 0)              {                 // Update all                 // multiples of p                 for (int i = p * 2;                           i < MAX; i += p)                     isPrime[i] = 1;             }         }         return -1;     }          static int findSubset(int []a, int n)     {         int cnt1 = 0;                // Counting no. of          // ones in the array         for (int i = 0; i < n; i++)              if (a[i] == 1)                 cnt1++;                // Case-I: count of          // ones(1s) > 0 and          // an integer > 1 is          // present in the array         if (cnt1 > 0)          {             for (int i = 0; i < n; i++)              {                 // Find a[i], where                 // a[i] + 1 is prime.                 if ((a[i] != 1) &&                      (isPrime[a[i] + 1] == 0))                  {                     System.out.println(cnt1 + 1);                            // Print all                      // the ones(1s).                     for (int j = 0;                              j < cnt1; j++)                                     System.out.print(1 + " ");                     System.out.println(a[i]); // print a[i].                     return 0;                 }             }         }                // Case-II: array contains         // only ones(1s)         if (cnt1 >= 2)          {             System.out.println(cnt1);                    // Print all ones(1s).             for (int i = 0;                       i < cnt1; i++)                  System.out.print(1 + " ");                    System.out.println();             return 0;         }                // Case-III: array does         // not contain 1s         for (int i = 0; i < n; i++)          {             for (int j = i + 1;                       j < n; j++)              {                 // Find a pair of integers                 // whose sum is prime                 if (isPrime[a[i] + a[j]] == 0)                  {                     System.out.println(2);                     System.out.println(a[i] +                                  " " + a[j]);                     return 0;                 }             }         }                // Array contains only          // a single element.         System.out.println(-1);         return -1;     }            // Driver Code     public static void main(String args[])     {         sieve();         int []A = new int[]{ 2, 1, 1 };         int n = A.length;         findSubset(A, n);     } }    // This code is contributed by  // Manish Shaw(manishshaw1)

Python3

 # Python3 program to find a subset in which  # sum of every pair in it is a prime import math as mt    MAX = 100001    isPrime = [0 for i in range(MAX)]    def sieve():        for p in range(2, mt.ceil(mt.sqrt(MAX))):                     # If isPrime[p] is not changed,          # then it is a prime         if (isPrime[p] == 0) :                            # Update all multiples of p             for i in range(2 * p, MAX, p):                 isPrime[i] = 1    def findSubset(a, n):        cnt1 = 0        # Counting no.of ones in the array     for i in range(n):          if (a[i] == 1):             cnt1+=1        # Case-I: count of ones(1s) > 0 and      # an integer > 1 is present in the array     if (cnt1 > 0):            for i in range(n):                # Find a[i], where a[i] + 1 is prime.             if ((a[i] != 1) and                 (isPrime[a[i] + 1] == 0)):                    print(cnt1 + 1)                    # Print all the ones(1s).                 for j in range(cnt1):                     print("1", end = " ")                    print(a[i])                 return 0        # Case-II: array contains only ones(1s)     if (cnt1 >= 2):            print(cnt1)            # Print all ones(1s).         for i in range(cnt1):             print("1", end = " ")            print("\n")         return 0            # Case-III: array does not contain 1s     for i in range(n):         for j in range(i + 1, n):                            # Find a pair of integers whose              # sum is prime             if (isPrime[a[i] + a[j]] == 0):                 print(2)                 print(a[i], " ", a[j])        # Array contains only a single element.     print(-1)    # Driver Code sieve() A =[ 2, 1, 1] n =len(A) findSubset(A, n)    # This code is contributed  # by Mohit kumar 29

C#

 // C# program to find a subset  // in which sum of every pair  // in it is a prime using System;    class GFG {     static int MAX = 100001;            static int []isPrime = new int[MAX];            static int sieve()     {         for (int p = 2;                   p * p < MAX; p++)          {             // If isPrime[p] is              // not changed, then             // it is a prime             if (isPrime[p] == 0)              {                 // Update all                 // multiples of p                 for (int i = p * 2;                           i < MAX; i += p)                     isPrime[i] = 1;             }         }         return -1;     }          static int findSubset(int []a, int n)     {         int cnt1 = 0;                // Counting no. of          // ones in the array         for (int i = 0; i < n; i++)              if (a[i] == 1)                 cnt1++;                // Case-I: count of          // ones(1s) > 0 and          // an integer > 1 is          // present in the array         if (cnt1 > 0)          {             for (int i = 0; i < n; i++)              {                 // Find a[i], where                 // a[i] + 1 is prime.                 if ((a[i] != 1) &&                      (isPrime[a[i] + 1] == 0))                  {                     Console.WriteLine(cnt1 + 1);                            // Print all the ones(1s).                     for (int j = 0; j < cnt1; j++)                                     Console.Write(1 + " ");                     Console.WriteLine(a[i]); // print a[i].                     return 0;                 }             }         }                // Case-II: array contains         // only ones(1s)         if (cnt1 >= 2)          {             Console.WriteLine(cnt1);                    // Print all ones(1s).             for (int i = 0; i < cnt1; i++)                  Console.Write(1 + " ");                    Console.WriteLine();             return 0;         }                // Case-III: array does         // not contain 1s         for (int i = 0; i < n; i++)          {             for (int j = i + 1; j < n; j++)              {                 // Find a pair of integers                 // whose sum is prime                 if (isPrime[a[i] + a[j]] == 0)                  {                     Console.WriteLine(2);                     Console.WriteLine(a[i] + " " + a[j]);                     return 0;                 }             }         }                // Array contains only          // a single element.         Console.WriteLine(-1);         return -1;     }            // Driver Code     static void Main()     {         sieve();         int []A = new int[]{ 2, 1, 1 };         int n = A.Length;         findSubset(A, n);     } }    // This code is contributed by  // Manish Shaw(manishshaw1)

PHP

 0 and      // an integer > 1 is      // present in the array     if (\$cnt1 > 0)      {         for (\$i = 0; \$i < \$n; \$i++)          {             // Find a[i], where              // a[i] + 1 is prime.             if ((\$a[\$i] != 1) and                  (\$isPrime[\$a[\$i] + 1] == 0))              {                 echo ((\$cnt1 + 1) . "\n");                    // Pr\$all the ones(1s).                 for (\$j = 0;                      \$j < \$cnt1; \$j++)                  {                     echo ("1 ");                 }                 echo (\$a[\$i] . "\n");                 return 0;             }         }     }        // Case-II: array contains      // only ones(1s)     if (\$cnt1 >= 2)      {         echo (cnt1 . "\n");            // Pr\$all ones(1s).         for (\$i = 0;               \$i < \$cnt1; \$i++)              echo ("1 ");            echo ("\n");         return 0;     }             // Case-III: array does      // not contain 1s     for (\$i = 0; \$i < \$n; \$i++)      {         for (\$j = \$i + 1;              \$j < \$n; \$j++)          {             // Find a pair of integers             // whose sum is prime             if (\$isPrime[\$a[\$i] +                  \$a[\$j]] == 0)              {                 echo (2 . "\n");                 echo (\$a[\$i] . " " .                        \$a[\$j] . "\n");                 return 0;             }         }     }        // Array contains only     // a single element.     echo (-1 . "\n"); }    // Driver Code sieve(); \$A = array(2, 1, 1); \$n = count(\$A); findSubset(\$A, \$n);    // This code is contributed by  // Manish Shaw(manishshaw1) ?>

Output :

3
1 1 2

Time Complexity : O(n2)

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