Bitwise XOR of elements having odd frequency

Given an array arr[] of N elements, the task is to find the XOR of the elements which appear odd number of times in the array.

Examples:

Input: arr[] = {1, 2, 1, 3, 3, 4, 2, 3, 1}
Output: 6
Elements with odd frequencies are 1, 3 and 4.
And (1 ^ 3 ^ 4) = 6



Input: arr[] = {2, 2, 7, 8, 7}
Output: 8

Naive Approach: Traverse the array and store the frequencies of all the elements in a unordered_map. Now, calculate the XOR of elements having odd frequency using the map created in the previous step.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the xor of
// elements having odd frequency
int xorOdd(int arr[], int n)
{
    // To store the frequency
    // of all the elements
    unordered_map<int, int> m;
  
    // Update the map with the
    // frequency of the elements
    for (int i = 0; i < n; i++)
        m[arr[i]]++;
  
    // To store the XOR of the elements
    // appearing odd number of
    // times in the array
    int xorArr = 0;
  
    // Traverse the map using an iterator
    for (auto it = m.begin(); it != m.end(); it++) {
  
        // Check for odd frequency
        // and update the xor
        if ((it->second) & 1) {
            xorArr ^= it->first;
        }
    }
  
    return xorArr;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 1, 3, 3, 4, 2, 3, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << xorOdd(arr, n);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG
{
      
// Function to return the xor of
// elements having odd frequency
static int xorOdd(int arr[], int n)
{
    // To store the frequency
    // of all the elements
    HashMap<Integer, 
            Integer> mp = new HashMap<Integer,
                                      Integer>();
  
    // Update the map with the
    // frequency of the elements
    for (int i = 0 ; i < n; i++)
    {
        if(mp.containsKey(arr[i]))
        {
            mp.put(arr[i], mp.get(arr[i]) + 1);
        }
        else
        {
            mp.put(arr[i], 1);
        }
    }
      
    // To store the XOR of the elements
    // appearing odd number of
    // times in the array
    int xorArr = 0;
  
    // Traverse the map using an iterator
    for (Map.Entry<Integer,
                   Integer> it : mp.entrySet()) 
    {
        // Check for odd frequency
        // and update the xor
        if (((it.getValue()) % 2) ==1)
        {
            xorArr ^= it.getKey();
        }
    }
    return xorArr;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 1, 3, 3, 4, 2, 3, 1 };
    int n = arr.length;
  
    System.out.println(xorOdd(arr, n));
}
}
  
// This code contributed by PrinciRaj1992

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Python3

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# Python3 implementation of the approach
  
# Function to return the xor of 
# elements having odd frequency 
def xorOdd(arr, n) : 
  
    # To store the frequency 
    # of all the elements 
    m = dict.fromkeys(arr, 0); 
  
    # Update the map with the 
    # frequency of the elements 
    for i in range(n) :
        m[arr[i]] += 1
  
    # To store the XOR of the elements 
    # appearing odd number of 
    # times in the array 
    xorArr = 0
  
    # Traverse the map using an iterator 
    for key,value in m.items() :
  
        # Check for odd frequency 
        # and update the xor 
        if (value & 1) :
            xorArr ^= key; 
  
    return xorArr; 
  
# Driver code 
if __name__ == "__main__"
  
    arr = [ 1, 2, 1, 3, 3, 4, 2, 3, 1 ]; 
    n = len(arr); 
  
    print(xorOdd(arr, n)); 
  
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach
using System;
using System.Collections.Generic;                 
      
class GFG
{
      
// Function to return the xor of
// elements having odd frequency
static int xorOdd(int []arr, int n)
{
    // To store the frequency
    // of all the elements
    Dictionary<int
               int> mp = new Dictionary<int
                                        int>();
  
    // Update the map with the
    // frequency of the elements
    for (int i = 0 ; i < n; i++)
    {
        if(mp.ContainsKey(arr[i]))
        {
            mp[arr[i]] = mp[arr[i]] + 1;
        }
        else
        {
            mp.Add(arr[i], 1);
        }
    }
      
    // To store the XOR of the elements
    // appearing odd number of
    // times in the array
    int xorArr = 0;
  
    // Traverse the map using an iterator
    foreach(KeyValuePair<int, int> it in mp) 
    {
        // Check for odd frequency
        // and update the xor
        if (((it.Value) % 2) == 1)
        {
            xorArr ^= it.Key;
        }
    }
    return xorArr;
}
  
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 1, 3, 3, 4, 2, 3, 1 };
    int n = arr.Length;
  
    Console.WriteLine(xorOdd(arr, n));
    }
}
  
// This code is contributed by Princi Singh

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Output:

6

This solution takes O(n) time and O(n) space.

Efficient Approach:

This approach uses two important properties of XOR – a ^ a = 0 and 0 ^ a = a. Take XOR of all the elements in the array. The result will be the XOR of numbers that appears odd number of times since elements appearing even number of times eventually cancel out each other.

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#include<bits/stdc++.h>
using namespace std;
  
int xorOdd(int arr[], int n) {
    // initialise result as 0
    int result = 0;
  
    // take XOR of all elements
    for (int i = 0; i < n; ++i) {
        result ^= arr[i];
    }
      
     // return result
    return result;
}
  
// Driver code
int main() {
    int arr[] = { 1, 2, 1, 3, 3, 4, 2, 3, 1 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    
    cout << xorOdd(arr, n); 
    
    return 0; 
}

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Output:

6

This solution take O(n) time and O(1) space.

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