# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation of the approach ` `#include ` `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); ` ` `  `    ``cout << xorOdd(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// 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 mp = ``new` `HashMap(); ` ` `  `    ``// 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 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 `

## Python3

 `# 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 `

## C#

 `// 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 `

Output:

```6
```

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