# Number of prime pairs in an array

Given an array. The task is to count the possible pairs which can be formed using the elements of the array where both of the elements in the pair are prime.
Note: Pairs such as (a, b) and (b, a) should not be considered different.

Examples:

```Input: arr[] = {1, 2, 3, 5, 7, 9}
Output: 6
From the given array, prime pairs are
(2, 3), (2, 5), (2, 7), (3, 5), (3, 7), (5, 7)

Input: arr[] = {1, 4, 5, 9, 11}
Output: 1
```

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

A naive approach is to count all possible pairs in the array and check if both the elements in the pair are prime.

An efficient approach is to count a number of primes in the array using Sieve of Eratosthenes. Let it be C. Now, total number of possible pairs is equal to C*(C-1)/2.

Below is the implementation of the above approach:

## C++

 `// CPP program to find count of ` `// prime pairs in given array. ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find count of prime pairs ` `int` `pairCount(``int` `arr[], ``int` `n) ` `{ ` `    ``// Find maximum value in the array ` `    ``int` `max_val = *max_element(arr, arr + n); ` ` `  `    ``// USE SIEVE TO FIND ALL PRIME NUMBERS LESS ` `    ``// THAN OR EQUAL TO max_val ` `    ``// Create a boolean array "prime[0..n]". A ` `    ``// value in prime[i] will finally be false ` `    ``// if i is Not a prime, else true. ` `    ``vector<``bool``> prime(max_val + 1, ``true``); ` ` `  `    ``// Remaining part of SIEVE ` `    ``prime = ``false``; ` `    ``prime = ``false``; ` `    ``for` `(``int` `p = 2; p * p <= max_val; p++) { ` ` `  `        ``// If prime[p] is not changed, then ` `        ``// it is a prime ` `        ``if` `(prime[p] == ``true``) { ` ` `  `            ``// Update all multiples of p ` `            ``for` `(``int` `i = p * 2; i <= max_val; i += p) ` `                ``prime[i] = ``false``; ` `        ``} ` `    ``} ` ` `  `    ``// Find all primes in arr[] ` `    ``int` `count = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``if` `(prime[arr[i]]) ` `            ``count++; ` ` `  `    ``// return the count of ` `    ``// possible prime pairs ` `    ``// Number of unique pairs ` `    ``// with N elements is N*(N-1)/2 ` `    ``return` `(count * (count - 1)) / 2; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3, 4, 5, 6, 7 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << pairCount(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find count of ` `// prime pairs in given array. ` `import` `java.util.*; ` ` `  `class` `GFG { ` ` `  `    ``// Function to find count of prime pairs ` `    ``static` `int` `pairCount(``int` `arr[], ``int` `n) ` `    ``{ ` ` `  `        ``// Find maximum value in the array ` `        ``int` `max_val = Arrays.stream(arr).max().getAsInt(); ` ` `  `        ``// USE SIEVE TO FIND ALL PRIME NUMBERS LESS ` `        ``// THAN OR EQUAL TO max_val ` `        ``// Create a boolean array "prime[0..n]". A ` `        ``// value in prime[i] will finally be false ` `        ``// if i is Not a prime, else true. ` `        ``Vector prime = ``new` `Vector<>(max_val + ``1``); ` `        ``for` `(``int` `i = ``0``; i < max_val + ``1``; i++) { ` `            ``prime.add(``true``); ` `        ``} ` ` `  `        ``// Remaining part of SIEVE ` `        ``prime.add(``0``, Boolean.FALSE); ` `        ``prime.add(``1``, Boolean.FALSE); ` `        ``for` `(``int` `p = ``2``; p * p <= max_val; p++) { ` ` `  `            ``// If prime[p] is not changed, then ` `            ``// it is a prime ` `            ``if` `(prime.get(p) == ``true``) { ` ` `  `                ``// Update all multiples of p ` `                ``for` `(``int` `i = p * ``2``; i <= max_val; i += p) { ` `                    ``prime.add(i, Boolean.FALSE); ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``// Find all primes in arr[] ` `        ``int` `count = ``0``; ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` `            ``if` `(prime.get(arr[i])) { ` `                ``count++; ` `            ``} ` `        ``} ` ` `  `        ``// return the count of ` `        ``// possible prime pairs ` `        ``// Number of unique pairs ` `        ``// with N elements is N*(N-1)/2 ` `        ``return` `(count * (count - ``1``)) / ``2``; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7` `}; ` `        ``int` `n = arr.length; ` `        ``System.out.println(pairCount(arr, n)); ` `    ``} ` `} ` ` `  `/* This code has been contributed  ` `by PrinciRaj1992*/`

## Python3

 `# Python 3 program to find count of ` `# prime pairs in given array. ` `from` `math ``import` `sqrt ` ` `  `# Function to find count of prime pairs ` `def` `pairCount(arr, n): ` `     `  `    ``# Find maximum value in the array ` `    ``max_val ``=` `arr[``0``] ` `    ``for` `i ``in` `range``(``len``(arr)): ` `        ``if``(arr[i] > max_val): ` `            ``max_val ``=` `arr[i] ` ` `  `    ``# USE SIEVE TO FIND ALL PRIME NUMBERS ` `    ``# LESS THAN OR EQUAL TO max_val ` `    ``# Create a boolean array "prime[0..n]".  ` `    ``# A value in prime[i] will finally be  ` `    ``# false if i is Not a prime, else true. ` `    ``prime ``=` `[``True` `for` `i ``in` `range``(max_val ``+` `1``)] ` ` `  `    ``# Remaining part of SIEVE ` `    ``prime[``0``] ``=` `False` `    ``prime[``1``] ``=` `False` `    ``k ``=` `int``(sqrt(max_val)) ``+` `1` `    ``for` `p ``in` `range``(``2``, k, ``1``): ` `         `  `        ``# If prime[p] is not changed,  ` `        ``# then it is a prime ` `        ``if` `(prime[p] ``=``=` `True``): ` `             `  `            ``# Update all multiples of p ` `            ``for` `i ``in` `range``(p ``*` `2``, max_val ``+` `1``, p): ` `                ``prime[i] ``=` `False` `     `  `    ``# Find all primes in arr[] ` `    ``count ``=` `0` `    ``for` `i ``in` `range``(``0``, n, ``1``): ` `        ``if` `(prime[arr[i]]): ` `                ``count ``+``=` `1` ` `  `    ``# return the count of possible prime  ` `    ``# pairs. Number of unique pairs ` `    ``# with N elements is N*(N-1)/2 ` `    ``return` `(count ``*` `(count ``-` `1``)) ``/` `2` ` `  `# Driver code ` `if` `__name__ ``=``=``'__main__'``: ` `    ``arr ``=` `[``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``]  ` `    ``n ``=` `len``(arr) ` `    ``print``(``int``(pairCount(arr, n))) ` `     `  `# This code is contributed by ` `# Shahshank_Sharma `

## C#

 `// C# program to find count of ` `// prime pairs in given array. ` ` `  `using` `System; ` `using` `System.Linq; ` ` `  `class` `GFG { ` ` `  `    ``// Function to find count of prime pairs ` `    ``static` `int` `pairCount(``int``[] arr, ``int` `n) ` `    ``{ ` ` `  `        ``// Find maximum value in the array ` `        ``int` `max_val = arr.Max(); ` ` `  `        ``// USE SIEVE TO FIND ALL PRIME NUMBERS LESS ` `        ``// THAN OR EQUAL TO max_val ` `        ``// Create a boolean array "prime[0..n]". A ` `        ``// value in prime[i] will finally be false ` `        ``// if i is Not a prime, else true. ` `        ``bool``[] prime = ``new` `bool``[max_val + 1]; ` `        ``for` `(``int` `i = 0; i < max_val + 1; i++) { ` `            ``prime[i] = ``true``; ` `        ``} ` ` `  `        ``// Remaining part of SIEVE ` `        ``prime = ``false``; ` `        ``prime = ``false``; ` `        ``for` `(``int` `p = 2; p * p <= max_val; p++) { ` ` `  `            ``// If prime[p] is not changed, then ` `            ``// it is a prime ` `            ``if` `(prime[p] == ``true``) { ` ` `  `                ``// Update all multiples of p ` `                ``for` `(``int` `i = p * 2; i <= max_val; i += p) { ` `                    ``prime[i] = ``false``; ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``// Find all primes in arr[] ` `        ``int` `count = 0; ` `        ``for` `(``int` `i = 0; i < n; i++) { ` `            ``if` `(prime[arr[i]]) { ` `                ``count++; ` `            ``} ` `        ``} ` ` `  `        ``// return the count of ` `        ``// possible prime pairs ` `        ``// Number of unique pairs ` `        ``// with N elements is N*(N-1)/2 ` `        ``return` `(count * (count - 1)) / 2; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int``[] arr = { 1, 2, 3, 4, 5, 6, 7 }; ` `        ``int` `n = arr.Length; ` `        ``Console.WriteLine(pairCount(arr, n)); ` `    ``} ` `} ` ` `  `// This code has been contributed by ihritik `

## PHP

 ` prime(max_val + 1, true);  ` `    ``\$prime` `= ``array_fill``(0, ``\$max_val` `+ 1, true); ` ` `  `    ``// Remaining part of SIEVE  ` `    ``\$prime`` = false;  ` `    ``\$prime`` = false;  ` `    ``for` `(``\$p` `= 2; ``\$p` `* ``\$p` `<= ``\$max_val``; ``\$p``++) ` `    ``{  ` ` `  `        ``// If prime[p] is not changed, ` `        ``// then it is a prime  ` `        ``if` `(``\$prime``[``\$p``] == true)  ` `        ``{  ` ` `  `            ``// Update all multiples of p  ` `            ``for` `(``\$i` `= ``\$p` `* 2; ` `                 ``\$i` `<= ``\$max_val``; ``\$i` `+= ``\$p``)  ` `                ``\$prime``[``\$i``] = false;  ` `        ``}  ` `    ``}  ` ` `  `    ``// Find all primes in arr[]  ` `    ``\$count` `= 0;  ` `    ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++)  ` `        ``if` `(``\$prime``[``\$arr``[``\$i``]])  ` `            ``\$count``++;  ` ` `  `    ``// return the count of  ` `    ``// possible prime pairs  ` `    ``// Number of unique pairs  ` `    ``// with N elements is N*(N-1)/2  ` `    ``return` `(``\$count` `* (``\$count` `- 1)) / 2;  ` `}  ` ` `  `// Driver code  ` `\$arr` `= ``array` `(1, 2, 3, 4, 5, 6, 7 );  ` `\$n` `= sizeof(``\$arr``);  ` `echo` `pairCount(``\$arr``, ``\$n``);  ` ` `  `// This code is contributed by ajit... ` `?> `

Output:

```6
``` My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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