# Count of pairs in an Array whose sum is Prime

Given an array arr of size N elements, the task is to count the number of pairs of elements in the array whose sum is prime.
Examples:

Input: arr = {1, 2, 3, 4, 5}
Output:
Explanation: Pairs with sum as a prime number are: {1, 2}, {1, 4}, {2, 3}, {2, 5} and {3, 4}
Input: arr = {10, 20, 30, 40}
Output:
Explanation: No pair whose sum is a prime number exists.

Naive Approach:
Calculate the sum of every pair of elements in the array and check if that sum is a prime number or not.
Below code is the implementation of the above approach:

## C++

 // C++ code to count of pairs  // of elements in an array  // whose sum is prime  #include  using namespace std;     // Function to check whether a  // number is prime or not  bool isPrime(int num)  {      if (num == 0 || num == 1) {          return false;      }      for (int i = 2; i * i <= num; i++) {          if (num % i == 0) {              return false;          }      }      return true;  }     // Function to count total number of pairs  // of elements whose sum is prime  int numPairsWithPrimeSum(int* arr, int n)  {      int count = 0;      for (int i = 0; i < n; i++) {          for (int j = i + 1; j < n; j++) {              int sum = arr[i] + arr[j];              if (isPrime(sum)) {                  count++;              }          }      }      return count;  }     // Driver Code  int main()  {      int arr[] = { 1, 2, 3, 4, 5 };      int n = sizeof(arr) / sizeof(arr);      cout << numPairsWithPrimeSum(arr, n);      return 0;  }

## Java

 // Java code to find number of pairs of  // elements in an array whose sum is prime  import java.io.*;  import java.util.*;     class GFG {         // Function to check whether a number      // is prime or not      public static boolean isPrime(int num)      {          if (num == 0 || num == 1) {              return false;          }          for (int i = 2; i * i <= num; i++) {              if (num % i == 0) {                  return false;              }          }          return true;      }         // Function to count total number of pairs      // of elements whose sum is prime      public static int numPairsWithPrimeSum(          int[] arr, int n)      {          int count = 0;          for (int i = 0; i < n; i++) {              for (int j = i + 1; j < n; j++) {                  int sum = arr[i] + arr[j];                  if (isPrime(sum)) {                      count++;                  }              }          }          return count;      }         // Driver code      public static void main(String[] args)      {          int[] arr = { 1, 2, 3, 4, 5 };          int n = arr.length;          System.out.println(              numPairsWithPrimeSum(arr, n));      }  }

## Python3

 # Python3 code to find number of pairs of   # elements in an array whose sum is prime   import math     # Function to check whether a  # number is prime or not  def isPrime(num):             sq = int(math.ceil(math.sqrt(num)))         if num == 0 or num == 1:          return False        for i in range(2, sq + 1):          if num % i == 0:              return False        return True    # Function to count total number of pairs  # of elements whose sum is prime  def numPairsWithPrimeSum(arr, n):             count = 0            for i in range(n):          for j in range(i + 1, n):              sum = arr[i] + arr[j]                             if isPrime(sum):                  count += 1                        return count     # Driver Code  arr = [ 1, 2, 3, 4, 5 ]  n = len(arr)     print(numPairsWithPrimeSum(arr, n))     # This code is contributed by grand_master

## C#

 // C# code to find number of pairs of  // elements in an array whose sum is prime  using System;  class GFG{     // Function to check whether a number  // is prime or not  public static bool isPrime(int num)  {      if (num == 0 || num == 1)       {          return false;      }      for (int i = 2; i * i <= num; i++)       {          if (num % i == 0)          {              return false;          }      }      return true;  }     // Function to count total number of pairs  // of elements whose sum is prime  public static int numPairsWithPrimeSum(int[] arr,                                         int n)  {      int count = 0;      for (int i = 0; i < n; i++)       {          for (int j = i + 1; j < n; j++)           {              int sum = arr[i] + arr[j];              if (isPrime(sum))              {                  count++;              }          }      }      return count;  }     // Driver code  public static void Main()  {      int[] arr = { 1, 2, 3, 4, 5 };      int n = arr.Length;      Console.Write(numPairsWithPrimeSum(arr, n));  }  }     // This code is contributed by Nidhi_Biet

Output:

5


Time Complexity: Efficient Approach:
Precompute and store the primes by using Sieve of Eratosthenes. Now, for every pair of elements, check whether their sum is prime or not.
Below code is the implementation of the above approach:

## C++

 // C++ code to find number of pairs  // of elements in an array whose  // sum is prime  #include  using namespace std;     // Function for Sieve Of Eratosthenes  bool* sieveOfEratosthenes(int N)  {      bool* isPrime = new bool[N + 1];      for (int i = 0; i < N + 1; i++) {          isPrime[i] = true;      }      isPrime = false;      isPrime = false;      for (int i = 2; i * i <= N; i++) {          if (isPrime[i] == true) {              int j = 2;              while (i * j <= N) {                  isPrime[i * j] = false;                  j++;              }          }      }      return isPrime;  }     // Function to count total number of pairs  // of elements whose sum is prime  int numPairsWithPrimeSum(int* arr, int n)  {      int N = 2 * 1000000;      bool* isPrime = sieveOfEratosthenes(N);      int count = 0;      for (int i = 0; i < n; i++) {          for (int j = i + 1; j < n; j++) {              int sum = arr[i] + arr[j];              if (isPrime[sum]) {                  count++;              }          }      }      return count;  }     // Driver Code  int main()  {      int arr[] = { 1, 2, 3, 4, 5 };      int n = sizeof(arr) / sizeof(arr);      cout << numPairsWithPrimeSum(arr, n);      return 0;  }

## Java

 // Java code to find number of pairs of  // elements in an array whose sum is prime  import java.io.*;  import java.util.*;     class GFG {      // Function for Sieve Of Eratosthenes      public static boolean[] sieveOfEratosthenes(int N)      {          boolean[] isPrime = new boolean[N + 1];          for (int i = 0; i < N + 1; i++) {              isPrime[i] = true;          }          isPrime = false;          isPrime = false;          for (int i = 2; i * i <= N; i++) {              if (isPrime[i] == true) {                  int j = 2;                  while (i * j <= N) {                      isPrime[i * j] = false;                      j++;                  }              }          }          return isPrime;      }         // Function to count total number of pairs      // of elements whose sum is prime      public static int numPairsWithPrimeSum(          int[] arr, int n)      {          int N = 2 * 1000000;          boolean[] isPrime = sieveOfEratosthenes(N);          int count = 0;          for (int i = 0; i < n; i++) {              for (int j = i + 1; j < n; j++) {                  int sum = arr[i] + arr[j];                  if (isPrime[sum]) {                      count++;                  }              }          }          return count;      }         // Driver code      public static void main(String[] args)      {          int[] arr = { 1, 2, 3, 4, 5 };          int n = arr.length;          System.out.println(              numPairsWithPrimeSum(arr, n));      }  }

## C#

 // C# code to find number of pairs of  // elements in an array whose sum is prime  using System;     class GFG{         // Function for Sieve Of Eratosthenes  public static bool[] sieveOfEratosthenes(int N)  {      bool[] isPrime = new bool[N + 1];      for (int i = 0; i < N + 1; i++)      {          isPrime[i] = true;      }      isPrime = false;      isPrime = false;      for (int i = 2; i * i <= N; i++)       {          if (isPrime[i] == true)          {              int j = 2;              while (i * j <= N)              {                  isPrime[i * j] = false;                  j++;              }          }      }      return isPrime;  }     // Function to count total number of pairs  // of elements whose sum is prime  public static int numPairsWithPrimeSum(int[] arr,                                          int n)  {      int N = 2 * 1000000;      bool[] isPrime = sieveOfEratosthenes(N);      int count = 0;      for (int i = 0; i < n; i++)      {          for (int j = i + 1; j < n; j++)           {              int sum = arr[i] + arr[j];              if (isPrime[sum])              {                  count++;              }          }      }      return count;  }     // Driver code  public static void Main(String[] args)  {      int[] arr = { 1, 2, 3, 4, 5 };      int n = arr.Length;      Console.WriteLine(numPairsWithPrimeSum(arr, n));  }  }     // This code is contributed by 29AjayKumar

Output:

5


Time complexity: O(N^2) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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