// C++ program to Count all pair with given XOR // value x #include   using namespace std; // Returns count of pairs in arr[0..n-1] with XOR // value equals to x. int xorPairCount(int arr[] int n int x) {  int result = 0; // Initialize result  // create empty set that stores the visiting   // element of array.   // Refer below post for details of unordered_set  // https://www.geeksforgeeks.org/cpp/unordered_set-in-cpp-stl/  unordered_set<int> s;  for (int i=0; i<n ; i++)  {  // If there exist an element in set s  // with XOR equals to x^arr[i] that means  // there exist an element such that the  // XOR of element with arr[i] is equal to  // x then increment count.  if (s.find(x^arr[i]) != s.end())  result++;  // Make element visited  s.insert(arr[i]);  }  // return total count of pairs with XOR equal to x  return result; } // driver program int main() {  int arr[] = {5  4 10 15 7 6};  int n = sizeof(arr)/sizeof(arr[0]);  int x = 5;  cout << 'Count of pairs with given XOR = '  << xorPairCount(arr n x);  return 0; } 

// Java program to Count all pair with  // given XOR value x  import java.util.*; class GFG  {  // Returns count of pairs in arr[0..n-1] with XOR   // value equals to x.   static int xorPairCount(int arr[] int n int x)   {  int result = 0; // Initialize result   // create empty set that stores the visiting   // element of array.   // Refer below post for details of unordered_set   // https://www.geeksforgeeks.org/cpp/unordered_set-in-cpp-stl/   HashSet<Integer> s = new HashSet<Integer>();  for (int i = 0; i < n; i++)   {  // If there exist an element in set s   // with XOR equals to x^arr[i] that means   // there exist an element such that the   // XOR of element with arr[i] is equal to   // x then increment count.   if (s.contains(x ^ arr[i]) &&   (x ^ arr[i]) == (int) s.toArray()[s.size() - 1])   {  result++;  }  // Make element visited   s.add(arr[i]);  }  // return total count of   // pairs with XOR equal to x   return result;  }  // Driver code   public static void main(String[] args)   {  int arr[] = {5 4 10 15 7 6};  int n = arr.length;  int x = 5;  System.out.print('Count of pairs with given XOR = '  + xorPairCount(arr n x));  } } // This code contributed by Rajput-Ji 

# Python3 program to count all the pair  # with given xor # Returns count of pairs in arr[0..n-1]  # with XOR value equals to x. def xorPairCount(arr n x): result = 0 # Initialize result # create empty set that stores the  # visiting element of array.  s = set() for i in range(0 n): # If there exist an element in set s # with XOR equals to x^arr[i] that  # means there exist an element such  # that the XOR of element with arr[i]  # is equal to x then increment count. if(x ^ arr[i] in s): result = result + 1 # Make element visited s.add(arr[i]) return result # Driver Code if __name__ == '__main__': arr = [5 4 10 15 7 6] n = len(arr) x = 5 print('Count of pair with given XOR = ' + str(xorPairCount(arr n x))) # This code is contributed by Anubhav Natani 

// C# program to Count all pair with  // given XOR value x  using System; using System.Collections.Generic; class GFG  {  // Returns count of pairs in arr[0..n-1] with XOR   // value equals to x.   static int xorPairCount(int []arr int n int x)   {  int result = 0; // Initialize result   // create empty set that stores the visiting   // element of array.   // Refer below post for details of unordered_set   // https://www.geeksforgeeks.org/cpp/unordered_set-in-cpp-stl/   HashSet<int> s = new HashSet<int>();  for (int i = 0; i < n; i++)   {  // If there exist an element in set s   // with XOR equals to x^arr[i] that means   // there exist an element such that the   // XOR of element with arr[i] is equal to   // x then increment count.   if (s.Contains(x ^ arr[i]))   {  result++;  }  // Make element visited   s.Add(arr[i]);  }  // return total count of   // pairs with XOR equal to x   return result;  }  // Driver code   public static void Main()   {  int []arr = {5 4 10 15 7 6};  int n = arr.Length;  int x = 5;  Console.WriteLine('Count of pairs with given XOR = '  + xorPairCount(arr n x));  } } /* This code contributed by PrinciRaj1992 */ 

<script> // Javascript program to Count all pair with  // given XOR value x     // Returns count of pairs in arr[0..n-1] with XOR   // value equals to x.   function xorPairCount(arrnx)  {  let result = 0; // Initialize result     // create empty set that stores the visiting   // element of array.   // Refer below post for details of unordered_set   // https://www.geeksforgeeks.org/cpp/unordered_set-in-cpp-stl/   let s = new Set();    for (let i = 0; i < n; i++)   {  // If there exist an element in set s   // with XOR equals to x^arr[i] that means   // there exist an element such that the   // XOR of element with arr[i] is equal to   // x then increment count.   if (s.has(x ^ arr[i]) )   {  result++;  }    // Make element visited   s.add(arr[i]);  }    // return total count of   // pairs with XOR equal to x   return result;  }    // Driver code   let arr=[5 4 10 15 7 6];  let n = arr.length;  let x = 5;  document.write('Count of pairs with given XOR = '  + xorPairCount(arr n x));      // This code is contributed by unknown2108 </script> 

// C++ program to Count all pair with given XOR // value x #include   using namespace std; // Returns count of pairs in arr[0..n-1] with XOR // value equals to x. int xorPairCount(int arr[] int n int x) {  int result = 0; // Initialize result  // create empty map that stores counts of  // individual elements of array.  unordered_map<int int> m;  for (int i=0; i<n ; i++)  {  int curr_xor = x^arr[i];  // If there exist an element in map m  // with XOR equals to x^arr[i] that means  // there exist an element such that the  // XOR of element with arr[i] is equal to  // x then increment count.  if (m.find(curr_xor) != m.end())  result += m[curr_xor];  // Increment count of current element  m[arr[i]]++;  }  // return total count of pairs with XOR equal to x  return result; } // driver program int main() {  int arr[] = {2 5 2};  int n = sizeof(arr)/sizeof(arr[0]);  int x = 0;  cout << 'Count of pairs with given XOR = '  << xorPairCount(arr n x);  return 0; } 

// Java program to Count all pair with given XOR // value x import java.util.*; class GFG { // Returns count of pairs in arr[0..n-1] with XOR // value equals to x. static int xorPairCount(int arr[] int n int x) {  int result = 0; // Initialize result  // create empty map that stores counts of  // individual elements of array.  Map<IntegerInteger> m = new HashMap<>();  for (int i = 0; i < n ; i++)  {  int curr_xor = x^arr[i];  // If there exist an element in map m  // with XOR equals to x^arr[i] that means  // there exist an element such that the  // XOR of element with arr[i] is equal to  // x then increment count.  if (m.containsKey(curr_xor))  result += m.get(curr_xor);  // Increment count of current element  if(m.containsKey(arr[i]))  {  m.put(arr[i] m.get(arr[i]) + 1);  }  else{  m.put(arr[i] 1);  }  }  // return total count of pairs with XOR equal to x  return result; } // Driver code public static void main(String[] args)  {  int arr[] = {2 5 2};  int n = arr.length;  int x = 0;  System.out.println('Count of pairs with given XOR = '  + xorPairCount(arr n x)); } } // This code has been contributed by 29AjayKumar 

# Python3 program to Count all pair with  # given XOR value x # Returns count of pairs in arr[0..n-1]  # with XOR value equals to x. def xorPairCount(arr n x): result = 0 # Initialize result # create empty map that stores counts  # of individual elements of array. m = dict() for i in range(n): curr_xor = x ^ arr[i] # If there exist an element in map m # with XOR equals to x^arr[i] that # means there exist an element such that  # the XOR of element with arr[i] is equal  # to x then increment count. if (curr_xor in m.keys()): result += m[curr_xor] # Increment count of current element if arr[i] in m.keys(): m[arr[i]] += 1 else: m[arr[i]] = 1 # return total count of pairs # with XOR equal to x return result # Driver Code arr = [2 5 2] n = len(arr) x = 0 print('Count of pairs with given XOR = ' xorPairCount(arr n x)) # This code is contributed by Mohit Kumar 

// C# program to Count all pair with given XOR // value x using System;  using System.Collections.Generic; class GFG { // Returns count of pairs in arr[0..n-1] with XOR // value equals to x. static int xorPairCount(int []arr int n int x) {  int result = 0; // Initialize result  // create empty map that stores counts of  // individual elements of array.  Dictionary<intint> m = new Dictionary<intint>();  for (int i = 0; i < n ; i++)  {  int curr_xor = x^arr[i];  // If there exist an element in map m  // with XOR equals to x^arr[i] that means  // there exist an element such that the  // XOR of element with arr[i] is equal to  // x then increment count.  if (m.ContainsKey(curr_xor))  result += m[curr_xor];  // Increment count of current element  if(m.ContainsKey(arr[i]))  {  var val = m[arr[i]];  m.Remove(arr[i]);  m.Add(arr[i] val + 1);  }  else  {  m.Add(arr[i] 1);  }  }    // return total count of pairs with XOR equal to x  return result; } // Driver code public static void Main(String[] args)  {  int []arr = {2 5 2};  int n = arr.Length;  int x = 0;  Console.WriteLine('Count of pairs with given XOR = '  + xorPairCount(arr n x)); } } // This code has been contributed by 29AjayKumar 

<script> // Javascript program to Count all pair with given XOR // value x // Returns count of pairs in arr[0..n-1] with XOR // value equals to x. function xorPairCount(arr n x) {  let result = 0; // Initialize result    // create empty map that stores counts of  // individual elements of array.  let m = new Map();    for (let i = 0; i < n ; i++)  {  let curr_xor = x^arr[i];    // If there exist an element in map m  // with XOR equals to x^arr[i] that means  // there exist an element such that the  // XOR of element with arr[i] is equal to  // x then increment count.  if (m.has(curr_xor))  result += m.get(curr_xor);    // Increment count of current element  if(m.has(arr[i]))  {  m.set(arr[i] m.get(arr[i]) + 1);  }  else{  m.set(arr[i] 1);  }  }  // return total count of pairs with XOR equal to x  return result; } // Driver program   let arr = [2 5 2];  let n = arr.length;  let x = 0;  document.write('Count of pairs with given XOR = '  + xorPairCount(arr n x));   </script>