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XOR of all subarray XORs | Set 2
• Difficulty Level : Hard
• Last Updated : 24 Mar, 2021

Given an array of integers, we need to get total XOR of all subarray XORs where subarray XOR can be obtained by XORing all elements of it.
Examples :

```Input : arr[] = [3, 5, 2, 4, 6]
Output : 7
Total XOR of all subarray XORs is,
(3) ^ (5) ^ (2) ^ (4) ^ (6)
(3^5) ^ (5^2) ^ (2^4) ^ (4^6)
(3^5^2) ^ (5^2^4) ^ (2^4^6)
(3^5^2^4) ^ (5^2^4^6) ^
(3^5^2^4^6) = 7

Input : arr[] = {1, 2, 3, 4}
Output : 0
Total XOR of all subarray XORs is,
(1) ^ (2) ^ (3) ^ (4) ^
(1^2) ^ (2^3) ^ (3^4) ^
(1^2^3) ^ (2^3^4) ^
(1^2^3^4) = 0```

We have discussed a O(n) solution in below post.
XOR of all subarray XORs | Set 1
As discussed in above post, frequency of element at i-th index is given by (i+1)*(N-i), where N is the size of the array
There are 4 cases possible:
Case 1: i is odd, N is odd
Let i = 2k+1, N = 2m+1
freq[i] = ((2k+1)+1)*((2m+1)-(2k+1)) = 4(m-k)(k+1) = even
Case 2: i is odd, N is even
Let i = 2k+1, N = 2m
freq[i] = ((2k+1)+1)*((2m)-(2k+1)) = 2(k+1)(2m-2k-1) = even
Case 3: i is even, N is odd
Let i = 2k, N = 2m+1
freq[i] = ((2k)+1)*((2m+1)-(2k)) = 2k(2m-2k+1)+(2m-2k)+1 = odd
Case 4: i is even, N is even
Let i = 2k, N = 2m
freq[i] = ((2k)+1)*((2m)-(2k)) = 2(m-k)(2k+1) = even
From this, we can conclude that if total no.of elements in the array is even, then frequency of element at any position is even. So total XOR will be 0. And if total no. of elements are odd, then frequency of elements at even positions are odd and add positions are even. So we need to find only the XOR of elements at even positions.
Below is implementation of above idea :

## C++

 `// C++ program to get total xor of all subarray xors``#include ``using` `namespace` `std;` `// Returns XOR of all subarray xors``int` `getTotalXorOfSubarrayXors(``int` `arr[], ``int` `N)``{``    ``// if even number of terms are there, all``    ``// numbers will appear even number of times.``    ``// So result is 0.``    ``if` `(N % 2 == 0)``       ``return` `0;` `    ``// else initialize result by 0 as (a xor 0 = a)``    ``int` `res = 0;``    ``for` `(``int` `i = 0; i

## Java

 `// Java program to get total``// xor of all subarray xors``import` `java.io.*;` `class` `GFG``{``    ``// Returns XOR of all``    ``// subarray xors``    ``static` `int` `getTotalXorOfSubarrayXors(``int` `arr[],``                                         ``int` `N)``    ``{``        ` `    ``// if even number of terms are``    ``// there, all numbers will appear``    ``// even number of times. So result is 0.``    ``if` `(N % ``2` `== ``0``)``    ``return` `0``;` `    ``// else initialize result``    ``// by 0 as (a xor 0 = a)``    ``int` `res = ``0``;``    ``for` `(``int` `i = ``0``; i < N; i += ``2``)``        ``res ^= arr[i];` `    ``return` `res;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main (String[] args)``    ``{``    ``int` `arr[] = {``3``, ``5``, ``2``, ``4``, ``6``};``    ``int` `N = arr.length;``        ` `    ``System.out.println(``            ``getTotalXorOfSubarrayXors(arr, N));``    ``}``}` `// This code is contributed by ajit`

## Python 3

 `# Python 3 program to get total xor``# of all subarray xors` `# Returns XOR of all subarray xors``def` `getTotalXorOfSubarrayXors(arr, N):` `    ``# if even number of terms are there,``    ``# all numbers will appear even number``    ``# of times. So result is 0.``    ``if` `(N ``%` `2` `=``=` `0``):``        ``return` `0` `    ``# else initialize result by 0``    ``# as (a xor 0 = a)``    ``res ``=` `0``    ``for` `i ``in` `range``(``0``, N, ``2``):``        ``res ^``=` `arr[i]` `    ``return` `res` `# Driver code``if` `__name__ ``=``=` `"__main__"``:` `    ``arr ``=` `[``3``, ``5``, ``2``, ``4``, ``6``]``    ``N ``=` `len``(arr)``    ``print``(getTotalXorOfSubarrayXors(arr, N))` `# This code is contributed by ita_c`

## C#

 `// C# program to get total``// xor of all subarray xors``using` `System;` `class` `GFG``{``    ` `    ``// Returns XOR of all``    ``// subarray xors``    ``static` `int` `getTotalXorOfSubarrayXors(``int` `[]arr,``                                         ``int` `N)``    ``{``        ` `    ``// if even number of terms``    ``// are there, all numbers``    ``// will appear even number``    ``// of times. So result is 0.``    ``if` `(N % 2 == 0)``    ``return` `0;` `    ``// else initialize result``    ``// by 0 as (a xor 0 = a)``    ``int` `res = 0;``    ``for` `(``int` `i = 0; i < N; i += 2)``        ``res ^= arr[i];` `    ``return` `res;``    ``}``    ` `    ``// Driver Code``    ``static` `void` `Main()``    ``{``    ``int` `[]arr = {3, 5, 2, 4, 6};``    ``int` `N = arr.Length;``    ``Console.Write(getTotalXorOfSubarrayXors(arr, N));``}``}` `// This code is contributed by aj_36`

## PHP

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## Javascript

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Output:

`7`

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