Bitwise XOR of all unordered pairs from a given array

Given an array arr[] of size N, the task is to find the bitwise XOR of all possible unordered pairs of the given array.

Examples:

Input: arr[] = {1, 5, 3, 7}
Output: 0
Explanation:
All possible unordered pairs are (1, 5), (1, 3), (1, 7), (5, 3), (5, 7), (3, 7)
Bitwise XOR of all possible pairs are = 1 ^ 5 ^ 1 ^3 ^ 1 ^ 7 ^ 5 ^ 3 ^ 5^ 7 ^ 3 ^ 7 = 0
Therefore, the required output is 0.

Input: arr[] = {1, 2, 3, 4}
Output: 4

Naive approach: The idea is to traverse the array and generate all possible pairs of the given array. Finally, print the Bitwise XOR of each element present in these pairs of the given array. Follow the steps below to solve the problem:

• Initialize a variable, say totalXOR, to store Bitwise XOR of each element from these pairs.
• Traverse the given array and generate all possible pairs(arr[i], arr[j]) from the given array.
• For each pair (arr[i], arr[j]), update the value of totalXOR = (totalXOR ^ arr[i] ^ arr[j]).
• Finally, print the value of totalXOR.

Below is the implementation of the above approach:

C++

 // C++ program to implement// the above approach #include using namespace std; // Function to get bitwise XOR// of all possible pairs of// the given arrayint TotalXorPair(int arr[], int N){    // Stores bitwise XOR    // of all possible pairs    int totalXOR = 0;     // Generate all possible pairs    // and calculate bitwise XOR    // of all possible pairs    for (int i = 0; i < N; i++) {        for (int j = i + 1; j < N;             j++) {             // Calculate bitwise XOR            // of each pair            totalXOR ^= arr[i]                        ^ arr[j];        }    }    return totalXOR;} // Driver Codeint main(){    int arr[] = { 1, 2, 3, 4 };    int N = sizeof(arr) / sizeof(arr[0]);    cout << TotalXorPair(arr, N);}

Java

 // Java program to implement // the above approach class GFG{   // Function to get bitwise XOR // of all possible pairs of // the given array public static int TotalXorPair(int arr[],                                int N) {   // Stores bitwise XOR   // of all possible pairs   int totalXOR = 0;    // Generate all possible pairs   // and calculate bitwise XOR   // of all possible pairs   for (int i = 0; i < N; i++)   {     for (int j = i + 1; j < N; j++)     {       // Calculate bitwise XOR       // of each pair       totalXOR ^= arr[i] ^ arr[j];     }   }      return totalXOR; }  // Driver code    public static void main(String[] args) {  int arr[] = {1, 2, 3, 4};   int N = arr.length;  System.out.print(TotalXorPair(arr, N));}} // This code is contributed by divyeshrabadiya07

Python3

 # Python3 program to implement# the above approach # Function to get bitwise XOR# of all possible pairs of# the given arraydef TotalXorPair(arr, N):       # Stores bitwise XOR    # of all possible pairs    totalXOR = 0;     # Generate all possible pairs    # and calculate bitwise XOR    # of all possible pairs    for i in range(0, N):        for j in range(i + 1, N):                       # Calculate bitwise XOR            # of each pair            totalXOR ^= arr[i] ^ arr[j];     return totalXOR; # Driver codeif __name__ == '__main__':       arr = [1, 2, 3, 4];    N = len(arr);    print(TotalXorPair(arr, N)); # This code is contributed by shikhasingrajput

C#

 // C# program to implement // the above approach using System; class GFG{   // Function to get bitwise XOR // of all possible pairs of // the given array public static int TotalXorPair(int []arr,                                int N) {      // Stores bitwise XOR   // of all possible pairs   int totalXOR = 0;    // Generate all possible pairs   // and calculate bitwise XOR   // of all possible pairs   for(int i = 0; i < N; i++)   {     for(int j = i + 1; j < N; j++)     {              // Calculate bitwise XOR       // of each pair       totalXOR ^= arr[i] ^ arr[j];     }   }   return totalXOR; }  // Driver code    public static void Main(String[] args) {  int []arr = {1, 2, 3, 4};   int N = arr.Length;     Console.Write(TotalXorPair(arr, N));}} // This code is contributed by Princi Singh

Javascript



Output
4

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, follow the observations below:

Property of Bitwise XOR:
a ^ a ^ a …….( Even times ) = 0
a ^ a ^ a …….( Odd times ) = a

Each element of the given array occurs exactly (N – 1) times in all possible pairs.
Therefore, if N is even, then Bitwise XOR of all possible pairs are equal to bitwise XOR of all the array elements.
Otherwise, bitwise XOR of all possible pairs are equal to 0.

Follow the steps below to solve the problem:

• If N is odd then print 0.
• If N is even then print the value of bitwise XOR of all the elements of the given array.

Below is the implementation of the above approach

C++

 // C++ program to implement// the above approach #include using namespace std; // Function to get bitwise XOR// of all possible pairs of// the given arrayint TotalXorPair(int arr[], int N){    // Stores bitwise XOR    // of all possible pairs    int totalXOR = 0;     // Check if N is odd    if (N % 2 != 0) {        return 0;    }     // If N is even then calculate    // bitwise XOR of all elements    // of the given array.    for (int i = 0; i < N; i++) {        totalXOR ^= arr[i];    }    return totalXOR;} // Driver Codeint main(){    int arr[] = { 1, 2, 3, 4 };    int N = sizeof(arr) / sizeof(arr[0]);    cout << TotalXorPair(arr, N);}

Java

 // Java program to implement // the above approach class GFG{   // Function to get bitwise XOR // of all possible pairs of // the given array public static int TotalXorPair(int arr[],                                int N) {   // Stores bitwise XOR   // of all possible pairs   int totalXOR = 0;    // Check if N is odd  if( N % 2 != 0 )  {    return 0;  }   // If N is even then calculate  // bitwise XOR of all elements  // of the array  for(int i = 0; i < N; i++)   {     totalXOR ^= arr[i];   }    return totalXOR; }  // Driver code    public static void main(String[] args) {  int arr[] = {1, 2, 3, 4};   int N = arr.length;  System.out.print(TotalXorPair(arr, N));}} // This code is contributed by math_lover

Python3

 # Python3 program to implement# the above approach # Function to get bitwise XOR# of all possible pairs of# the given array def TotalXorPair(arr, N):     # Stores bitwise XOR    # of all possible pairs    totalXOR = 0     # Check if N is odd    if (N % 2 != 0):        return 0     # If N is even then calculate    # bitwise XOR of all elements    # of the given array.    for i in range(N):        totalXOR ^= arr[i]     return totalXOR # Driver codeif __name__ == '__main__':     arr = [ 1, 2, 3, 4 ]    N = len(arr)         print(TotalXorPair(arr, N)) # This code is contributed by Shivam Singh

C#

 // C# program to implement // the above approach using System; class GFG{   // Function to get bitwise XOR // of all possible pairs of // the given array static int TotalXorPair(int []arr,                         int N) {          // Stores bitwise XOR     // of all possible pairs     int totalXOR = 0;          // Check if N is odd    if (N % 2 != 0)    {        return 0;    }         // If N is even then calculate    // bitwise XOR of all elements    // of the array    for(int i = 0; i < N; i++)     {         totalXOR ^= arr[i];     }     return totalXOR; }  // Driver code    public static void Main(String[] args) {    int []arr = { 1, 2, 3, 4 };     int N = arr.Length;         Console.Write(TotalXorPair(arr, N));}} // This code is contributed by doreamon_

Javascript



Output
4

Time Complexity: O(N)
Auxiliary Space: O(1)

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