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Bitwise XOR of elements having odd frequency
• Last Updated : 27 Oct, 2020

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

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

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

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++

 `// 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[0]);`   `    ``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

```

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 an odd number of times since elements appearing even number of times eventually cancel out each other.

## C++

 `// C++ program to implement` `// the above approach` `#include` `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; ` `}`

## Java

 `// Java program to implement` `// the above approach` `import` `java.io.*;`   `class` `GFG{`   `static` `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` `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 is contributed by math_lover`

## Python3

 `# Python3 program to implement` `# the above approach` `def` `xorOdd(arr, n):` `  `  `    ``# Initialise result as 0` `    ``result ``=` `0`   `    ``# Take XOR of all elements` `    ``for` `i ``in` `range` `(n):` `        ``result ^``=` `arr[i]` `    `  `     ``# Return result` `    ``return` `result`   `# 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 Chitranayal`

Output:

```6

```

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

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