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++

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// C++ code to count of pairs
// of elements in an array
// whose sum is prime
#include <bits/stdc++.h>
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[0]);
    cout << numPairsWithPrimeSum(arr, n);
    return 0;
}

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Java

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// 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));
    }
}

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Python3

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# 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

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C#

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// 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

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Output: 



5

 

Time Complexity: 

O(N^2 * \sqrt{sum})

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++

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// C++ code to find number of pairs
// of elements in an array whose
// sum is prime
#include <bits/stdc++.h>
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[0] = false;
    isPrime[1] = 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[0]);
    cout << numPairsWithPrimeSum(arr, n);
    return 0;
}

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Java

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// 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[0] = false;
        isPrime[1] = 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));
    }
}

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C#

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// 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[0] = false;
    isPrime[1] = 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

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Output: 

5

 

Time complexity: O(N^2)
 

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