All pairs whose xor gives unique prime

Given an array arr[], the task is to count all the pairs whose xor gives the unique prime, i.e. no two pairs should give the same prime.

Examples:

Input: arr[] = {2, 3, 4, 5, 6, 7, 8, 9}
Output: 6
(2, 5), (2, 7), (2, 9), (4, 6), (4, 7) and (4, 9) are the only pairs whose XORs are primes i.e. 7, 5, 11, 2, 3 and 13 respectively.

Input: arr[] = {10, 12, 23, 45, 5, 6}
Output: 4

Approach: Iterating every possible pair and check whether the xor of the current pair is a prime. If its a prime then update count = count + 1 and also save the prime in an unordered_map in order to keep track of the repeating primes. Print the count in the end.

Below is the implementation of the above approach:

C++

 `// C++ implementation of above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function that returns true if n is prime ` `bool` `isPrime(``int` `n) ` `{ ` `    ``// Corner cases ` `    ``if` `(n <= 1) ` `        ``return` `false``; ` `    ``if` `(n <= 3) ` `        ``return` `true``; ` ` `  `    ``// This is checked so that we can skip ` `    ``// middle five numbers in below loop ` `    ``if` `(n % 2 == 0 || n % 3 == 0) ` `        ``return` `false``; ` ` `  `    ``for` `(``int` `i = 5; i * i <= n; i = i + 6) ` `        ``if` `(n % i == 0 || n % (i + 2) == 0) ` `            ``return` `false``; ` ` `  `    ``return` `true``; ` `} ` ` `  `// Function to return the count of valid pairs ` `int` `countPairs(``int` `a[], ``int` `n) ` `{ ` `    ``int` `count = 0; ` ` `  `    ``unordered_map<``int``, ``int``> m; ` ` `  `    ``for` `(``int` `i = 0; i < n - 1; i++) { ` `        ``for` `(``int` `j = i + 1; j < n; j++) { ` ` `  `            ``// If xor(a[i], a[j]) is prime and unique ` `            ``if` `(isPrime(a[i] ^ a[j]) && m[a[i] ^ a[j]] == 0) { ` `                ``m[(a[i] ^ a[j])]++; ` `                ``count++; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a[] = { 10, 12, 23, 45, 5, 6 }; ` ` `  `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` ` `  `    ``cout << countPairs(a, n); ` `} `

Java

 `// Java implementation of above approach ` `import` `java.util.*; ` `class` `solution ` `{ ` `  `  `// Function that returns true if n is prime ` `static` `boolean` `isPrime(``int` `n) ` `{ ` `    ``// Corner cases ` `    ``if` `(n <= ``1``) ` `        ``return` `false``; ` `    ``if` `(n <= ``3``) ` `        ``return` `true``; ` `  `  `    ``// This is checked so that we can skip ` `    ``// middle five numbers in below loop ` `    ``if` `(n % ``2` `== ``0` `|| n % ``3` `== ``0``) ` `        ``return` `false``; ` `  `  `    ``for` `(``int` `i = ``5``; i * i <= n; i = i + ``6``) ` `        ``if` `(n % i == ``0` `|| n % (i + ``2``) == ``0``) ` `            ``return` `false``; ` `  `  `    ``return` `true``; ` `} ` `  `  `// Function to return the count of valid pairs ` `static` `int` `countPairs(``int` `a[], ``int` `n) ` `{ ` `    ``int` `count = ``0``; ` `  `  `    ``Map m=``new` `HashMap< Integer,Integer>(); ` `  `  `    ``for` `(``int` `i = ``0``; i < n - ``1``; i++) { ` `        ``for` `(``int` `j = i + ``1``; j < n; j++) { ` `  `  `            ``// If xor(a[i], a[j]) is prime and unique ` `            ``if` `(isPrime(a[i] ^ a[j]) && m.get(a[i] ^ a[j]) == ``null``) { ` `                ``m.put((a[i] ^ a[j]),``1``); ` `                ``count++; ` `            ``} ` `        ``} ` `    ``} ` `  `  `    ``return` `count; ` `} ` `  `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `a[] = { ``10``, ``12``, ``23``, ``45``, ``5``, ``6` `}; ` `  `  `    ``int` `n = a.length; ` `  `  `    ``System.out.println(countPairs(a, n)); ` `} ` `} ` `//contributed by Arnab Kundu `

C#

 `// C# implementation of the approach  ` `using` `System ; ` `using` `System.Collections ; ` ` `  `class` `solution  ` `{  ` ` `  `    ``// Function that returns true if n is prime  ` `    ``static` `bool` `isPrime(``int` `n)  ` `    ``{  ` `        ``// Corner cases  ` `        ``if` `(n <= 1)  ` `            ``return` `false``;  ` `        ``if` `(n <= 3)  ` `            ``return` `true``;  ` `     `  `        ``// This is checked so that we can skip  ` `        ``// middle five numbers in below loop  ` `        ``if` `(n % 2 == 0 || n % 3 == 0)  ` `            ``return` `false``;  ` `     `  `        ``for` `(``int` `i = 5; i * i <= n; i = i + 6)  ` `            ``if` `(n % i == 0 || n % (i + 2) == 0)  ` `                ``return` `false``;  ` `     `  `        ``return` `true``;  ` `    ``}  ` `     `  `    ``// Function to return the count of valid pairs  ` `    ``static` `int` `countPairs(``int` `[]a, ``int` `n)  ` `    ``{  ` `        ``int` `count = 0;  ` `     `  `        ``Hashtable m=``new` `Hashtable();  ` `     `  `        ``for` `(``int` `i = 0; i < n - 1; i++) {  ` `            ``for` `(``int` `j = i + 1; j < n; j++) {  ` `     `  `                ``// If xor(a[i], a[j]) is prime and unique  ` `                ``if` `(isPrime(a[i] ^ a[j]) && m[a[i] ^ a[j]] == ``null``) ` `                ``{  ` `                    ``m.Add((a[i] ^ a[j]),1);  ` `                    ``count++;  ` `                ``}  ` `            ``}  ` `        ``}  ` `     `  `        ``return` `count;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``int` `[]a = { 10, 12, 23, 45, 5, 6 };  ` `     `  `        ``int` `n = a.Length;  ` `     `  `        ``Console.WriteLine(countPairs(a, n));  ` `    ``}  ` `    ``// This code is contributed by Ryuga ` `}  `

Output:

```4
```

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