Find even occurring elements in an array of limited range

Given an array that contains odd number of occurrences for all numbers except for a few elements which are present even number of times. Find the elements which have even occurrences in the array in O(n) time complexity and O(1) extra space.
Assume array contain elements in the range 0 to 63.

Examples :

Input: [9, 12, 23, 10, 12, 12, 15, 23, 14, 12, 15]
Output: 12, 23 and 15

Input: [1, 4, 7, 5, 9, 7, 3, 4, 6, 8, 3, 0, 3]
Output: 4 and 7

Input: [4, 4, 10, 10, 4, 4, 4, 4, 10, 10]
Output: 4 and 10

A simple method would be to traverse the array and store frequencies of its elements in a map. Later, print elements that have even count in the map. The solution takes O(n) time but requires extra space for storing frequencies. Below is an interesting method to solve this problem using bitwise operators.

This method assumes that long long integers are stored using 64 bits. The idea is to use XOR operator. We know that

1 XOR 1 = 0
1 XOR 0 = 1
0 XOR 1 = 1
0 XOR 0 = 0

Consider below input –

1, 4, 7, 5, 9, 7, 3, 4, 6, 8, 3, 0, 3

If we right shift 1 by value of each element of the array and take XOR of all the items, we will get below binary integer –

1101101011

Each 1 in the i’th index from the right represents odd occurrence of element i. And each 0 in the i’th index from the right represents even or non-occurrence of element i in the array.

0 is present in 2nd, 4th and 7th position in above binary number. But 2 is not present in our array. So our answer is 4 and 7.

Below is the implementation of above idea

C++

 // C++ Program to find the even occurring elements // in given array #include using namespace std;    // Function to find the even occurring elements // in given array void printRepeatingEven(int arr[], int n) {     long long _xor = 0L;     long long pos;        // do for each element of array     for( int i = 0; i < n; ++i)     {         // right pos 1 by value of current element         pos = 1 << arr[i];            // Toggle the bit everytime element gets repeated         _xor ^= pos;     }        // Traverse array again and use _xor to find even     // occuring elements     for (int i = 0; i < n; ++i)     {         // right shift 1 by value of current element         pos = 1 << arr[i];            // Each 0 in _xor represents an even occurrence         if (!(pos & _xor))         {             // print the even occurring numbers             cout << arr[i] << " ";                // set bit as 1 to avoid printing duplicates             _xor ^= pos;         }     } }    // Driver code int main() {     int arr[] = { 9, 12, 23, 10, 12, 12, 15, 23,                   14, 12, 15 };     int n = sizeof(arr) / sizeof(arr);        printRepeatingEven(arr, n);        return 0; }

Java

 // Java Program to find the even occurring // elements in given array  class GFG  {    // Function to find the even occurring  // elements in given array  static void printRepeatingEven(int arr[],                                int n) {     long _xor = 0L;     long pos;        // do for each element of array      for (int i = 0; i < n; ++i)      {         // right pos 1 by value of          // current element          pos = 1 << arr[i];            // Toggle the bit everytime          // element gets repeated          _xor ^= pos;     }        // Traverse array again and use _xor      // to find even occuring elements      for (int i = 0; i < n; ++i)     {         // right shift 1 by value of          // current element          pos = 1 << arr[i];            // Each 0 in _xor represents         // an even occurrence          if (!((pos & _xor)!=0))          {             // print the even occurring numbers              System.out.print(arr[i] + " ");                // set bit as 1 to avoid              // printing duplicates              _xor ^= pos;         }     } }    // Driver code  public static void main(String args[])  {     int arr[] = {9, 12, 23, 10, 12, 12,                  15, 23, 14, 12, 15};     int n = arr.length;        printRepeatingEven(arr, n); } }    // This code is contributed  // by 29AjayKumar

C#

 // C# Program to find the even occurring // elements in given array  using System;    class GFG  {    // Function to find the even occurring  // elements in given array  static void printRepeatingEven(int[] arr,                                int n) {     long _xor = 0L;     long pos;        // do for each element of array      for (int i = 0; i < n; ++i)      {         // right pos 1 by value of          // current element          pos = 1 << arr[i];            // Toggle the bit everytime          // element gets repeated          _xor ^= pos;     }        // Traverse array again and use _xor      // to find even occuring elements      for (int i = 0; i < n; ++i)     {         // right shift 1 by value of          // current element          pos = 1 << arr[i];            // Each 0 in _xor represents         // an even occurrence          if (!((pos & _xor) != 0))          {             // print the even occurring numbers              Console.Write(arr[i] + " ");                // set bit as 1 to avoid              // printing duplicates              _xor ^= pos;         }     } }    // Driver code  public static void Main()  {     int[] arr = {9, 12, 23, 10, 12, 12,                 15, 23, 14, 12, 15};     int n = arr.Length;        printRepeatingEven(arr, n); } }    // This code is contributed  // by Mukul singh



Python 3

 # Python 3 program to find the even  # occurring elements in given array    # Function to find the even occurring  # elements in given array def printRepeatingEven(arr, n):        axor = 0;        # do for each element of array     for i in range(0, n):                # right pos 1 by value of          # current element         pos = 1 << arr[i];            # Toggle the bit everytime          # element gets repeated         axor ^= pos;            # Traverse array again and use _xor      # to find even occuring elements     for i in range(0, n - 1):                # right shift 1 by value of          # current element         pos = 1 << arr[i];            # Each 0 in _xor represents an          # even occurrence         if (not(pos & axor)):                        # print the even occurring numbers             print(arr[i], end = " ");                # set bit as 1 to avoid printing              # duplicates             axor ^= pos;            # Driver code arr = [9, 12, 23, 10, 12, 12,            15, 23, 14, 12, 15 ]; n = len(arr) ; printRepeatingEven(arr, n);    # This code is contributed  # by Shivi_Aggarwal

Output :

12 23 15