Bitwise XOR of all unordered pairs from a given array - GeeksforGeeks

26 Jul, 2021
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: 0Explanation: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 = 0Therefore, 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++
Java
Python3
C#
Javascript
// C++ program to implement// the above approach #include <bits/stdc++.h>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 program to implement// the above approachclass GFG{   // Function to get bitwise XOR// of all possible pairs of// the given arraypublic 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 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# program to implement// the above approachusing System; class GFG{   // Function to get bitwise XOR// of all possible pairs of// the given arraypublic 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
<script> // Javascript program to implement// the above approach // Function to get bitwise XOR// of all possible pairs of// the given arrayfunction TotalXorPair(arr, N){         // Stores bitwise XOR    // of all possible pairs    let totalXOR = 0;         // Generate all possible pairs    // and calculate bitwise XOR    // of all possible pairs    for(let i = 0; i < N; i++)    {        for(let j = i + 1; j < N; j++)        {                         // Calculate bitwise XOR            // of each pair            totalXOR ^= arr[i] ^ arr[j];        }    }    return totalXOR;}  // Driver Codelet arr = [ 1, 2, 3, 4 ];let N = arr.length; document.write(TotalXorPair(arr, N)); // This code is contributed by target_2 </script>
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++
Java
Python3
C#
Javascript
// C++ program to implement// the above approach #include <bits/stdc++.h>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 program to implement// the above approachclass GFG{   // Function to get bitwise XOR// of all possible pairs of// the given arraypublic 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 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     # 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# program to implement// the above approachusing System; class GFG{   // Function to get bitwise XOR// of all possible pairs of// the given arraystatic 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_
<script> // Javascript program to implement// the above approach // Function to get bitwise XOR// of all possible pairs of// the given arrayfunction TotalXorPair(arr, N){    // Stores bitwise XOR    // of all possible pairs    let 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 (let i = 0; i < N; i++) {        totalXOR ^= arr[i];    }    return totalXOR;} // Driver Code    let arr = [ 1, 2, 3, 4 ];    let N = arr.length;    document.write(TotalXorPair(arr, N)); </script>
4
Time Complexity: O(N)Auxiliary Space: O(1)
SHIVAMSINGH67
divyeshrabadiya07
math_lover
princi singh
doreamon_
shikhasingrajput
target_2
subhammahato348
akshaysingh98088
Bitwise-XOR
Arrays
Bit Magic
Combinatorial
Greedy
Mathematical
Arrays
Greedy
Mathematical
Bit Magic
Combinatorial
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