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# Sum of XOR of all possible subsets

• Difficulty Level : Medium
• Last Updated : 07 Jul, 2022

Given an array arr[] of size n, we need to find the sum of all the values that come from XORing all the elements of the subsets.

```Input :  arr[] = {1, 5, 6}
Output : 28
Total Subsets = 23
1 = 1
5 = 5
6 = 6
1 ^ 5 = 4
1 ^ 6 = 7
5 ^ 6 = 3
1 ^ 5 ^ 6 = 2
0(empty subset)
Now SUM of all these XORs = 1 + 5 + 6 + 4 +
7 + 3 + 2 + 0
= 28

Input : arr[] = {1, 2}
Output : 6
```

A Naive approach is to take the XOR all possible combinations of array[] elements and then perform the summation of all values. Time complexity of this approach grows exponentially so it would not be better for a large value of n.

Implementation: Recursive Code for the Naive Approach

## C++

 `#include ` `using` `namespace` `std;` `int` `rec(``int` `i, ``int` `x, ``int` `arr[], ``int` `size)` `{` `    ``// return the current xor sum if we reach the end of` `    ``// array` `    ``if` `(i == size)` `        ``return` `x;` `    ``// first choice can be to include the i-th element in` `    ``// the subset and thus we take its xor` `    ``int` `choice1 = rec(i + 1, x ^ arr[i], arr, size);`   `    ``// second choice can be to include the i-th element in` `    ``// the subset and thus we take its xor` `    ``int` `choice2 = rec(i + 1, x, arr, size);`   `    ``// return sum of both the choices as we need to find the` `    ``// sum of xor of all subsets` `    ``return` `choice1 + choice2;` `}` `// Returns sum of XORs of all subsets` `int` `xorSum(``int` `arr[], ``int` `size)` `{` `    ``return` `rec(0, 0, arr, size);` `}` `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 1, 5, 6 };` `    ``int` `size = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << xorSum(arr, size);` `}`

## Java

 `// Java program to implement the approach` `class` `GFG {` `    ``static` `int` `rec(``int` `i, ``int` `x, ``int` `arr[], ``int` `size)` `    ``{` `      `  `        ``// return the current xor sum if we reach the end of` `        ``// array` `        ``if` `(i == size)` `            ``return` `x;` `        ``// first choice can be to include the i-th element` `        ``// in the subset and thus we take its xor` `        ``int` `choice1 = rec(i + ``1``, x ^ arr[i], arr, size);`   `        ``// second choice can be to include the i-th element` `        ``// in the subset and thus we take its xor` `        ``int` `choice2 = rec(i + ``1``, x, arr, size);`   `        ``// return sum of both the choices as we need to find` `        ``// the sum of xor of all subsets` `        ``return` `choice1 + choice2;` `    ``}`   `    ``// Returns sum of XORs of all subsets` `    ``static` `int` `xorSum(``int` `arr[], ``int` `size)` `    ``{` `        ``return` `rec(``0``, ``0``, arr, size);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `arr[] = { ``1``, ``5``, ``6` `};` `        ``int` `size = arr.length;` `      `  `          ``//Function call` `        ``System.out.println(xorSum(arr, size));` `    ``}` `}`   `// This code is contributed by phasing17`

## Python3

 `# Python3 program to implement the approach` `def` `rec(i, x, arr, size):`   `    ``# return the current xor sum if we reach the end of` `    ``# array` `    ``if` `(i ``=``=` `size):` `        ``return` `x` `      `  `    ``# first choice can be to include the i-th element in` `    ``# the subset and thus we take its xor` `    ``choice1 ``=` `rec(i ``+` `1``, x ^ arr[i], arr, size)`   `    ``# second choice can be to include the i-th element in` `    ``# the subset and thus we take its xor` `    ``choice2 ``=` `rec(i ``+` `1``, x, arr, size)`   `    ``# return sum of both the choices as we need to find the` `    ``# sum of xor of all subsets` `    ``return` `choice1 ``+` `choice2`   `# Returns sum of XORs of all subsets` `def` `xorSum(arr, size):` `    ``return` `rec(``0``, ``0``, arr, size)`   `# Driver code` `arr ``=` `[``1``, ``5``, ``6``]` `size ``=` `len``(arr)`   `# Function call` `print``(xorSum(arr, size))`   `# This code is contributed by phasing17`

## C#

 `// C# program to implement the approach`   `using` `System;`   `class` `GFG {` `    ``static` `int` `rec(``int` `i, ``int` `x, ``int``[] arr, ``int` `size)` `    ``{`   `        ``// return the current xor sum if we reach the end of` `        ``// array` `        ``if` `(i == size)` `            ``return` `x;` `        ``// first choice can be to include the i-th element` `        ``// in the subset and thus we take its xor` `        ``int` `choice1 = rec(i + 1, x ^ arr[i], arr, size);`   `        ``// second choice can be to include the i-th element` `        ``// in the subset and thus we take its xor` `        ``int` `choice2 = rec(i + 1, x, arr, size);`   `        ``// return sum of both the choices as we need to find` `        ``// the sum of xor of all subsets` `        ``return` `choice1 + choice2;` `    ``}`   `    ``// Returns sum of XORs of all subsets` `    ``static` `int` `xorSum(``int``[] arr, ``int` `size)` `    ``{` `        ``return` `rec(0, 0, arr, size);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main(``string``[] args)` `    ``{` `        ``int``[] arr = { 1, 5, 6 };` `        ``int` `size = arr.Length;`   `        ``// Function call` `        ``Console.WriteLine(xorSum(arr, size));` `    ``}` `}`   `// This code is contributed by phasing17`

## Javascript

 ``

Output

`28`

An Efficient approach is to find the pattern with respect to the property of XOR. Now again consider the subset in binary form like:

```    1 = 001
5 = 101
6 = 110
1 ^ 5 = 100
1 ^ 6 = 111
5 ^ 6 = 011
1^5^6 = 010```

So if we analyze all these binary numbers of the XORs, we can observe that set bit occurs at all the positions of i(0 to n-1) will exactly contribute to half of 2n. So we can easily impose these two conditions at each such position of i.

• If there is any value of arr[] that has set ith bit set, then exactly half of 2n subsets will be of the form, so they will contribute to 2n-1+i to the final sum.
• If there is no value of arr[] that ith bit set, then we can say that there will be no term in all subsets that have a ith bit set.

The proof of the above point is as follows:

Case 1:

• Lets assume there are k elements in the array with ith bit set and k is not zero.
• So, to have a subset with ith bit set in its xor, we need it to have odd number of elements with ith bit set.
• Number of ways to choose elements with ith bit not set = 2(n-k)
• Number of ways to choose elements with ith bit set = kC1 + kC3 + kC5 …. = 2(k-1)
• Total number of ways = 2(n-1)
• Thus, the contribution towards sum becomes, 2(n+i-1)

Case 2:

• f no element has ith bit set, i.e. k = 0, the contribution of ith bit towards total sum remains 0.
• Now the question boils down to check which position of element of the arr[] will be set or not. But here is some trick that we will not iterate for all elements one by one in spite of that we can simple take the OR of all such values and multiply with 2n-1

For example

```Take a OR of all arr[] elements, we get
= 1 | 5 | 6
= 001 | 101 | 110
= 111

Now to find final summation, we can write it down as:-
= 1*2n-1+2 + 1*2n-1+1 + 1*2n-1+0
= 2n-1 * (1*22 + 1*21 + 1*20 )
= 2n-1 * (1112)
= 2n-1 * 7

Put n = 3,  we get
= 28```

So at last for any value of n and array elements, we can simple say that the final sum will be 2n-1 times the bitwise OR of all the inputs.

## C++

 `// Below is C++ approach to finding the XOR_SUM` `#include` `using` `namespace` `std;`   `// Returns sum of XORs of all subsets` `int` `xorSum(``int` `arr[], ``int` `n)` `{` `    ``int` `bits = 0;`   `    ``// Finding bitwise OR of all elements` `    ``for` `(``int` `i=0; i < n; ++i)` `        ``bits |= arr[i];`   `    ``int` `ans = bits * ``pow``(2, n-1);`   `    ``return` `ans;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = {1, 5, 6};` `    ``int` `size = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << xorSum(arr, size);` `}`

## Java

 `// Java approach to finding the XOR_SUM` `class` `GFG {` `    `  `    ``// Returns sum of XORs of all subsets` `    ``static` `int` `xorSum(``int` `arr[], ``int` `n)` `    ``{` `        `  `        ``int` `bits = ``0``;` `    `  `        ``// Finding bitwise OR of all elements` `        ``for` `(``int` `i = ``0``; i < n; ++i)` `            ``bits |= arr[i];` `    `  `        ``int` `ans = bits * (``int``)Math.pow(``2``, n-``1``);` `    `  `        ``return` `ans;` `    ``}` `    `  `    ``// Driver method` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        `  `        ``int` `arr[] = {``1``, ``5``, ``6``};` `        ``int` `size = arr.length;` `        `  `        ``System.out.print(xorSum(arr, size));` `    ``}` `}`   `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python3 approach to finding the XOR_SUM`   `# Returns sum of XORs of all subsets` `def` `xorSum(arr, n):`   `    ``bits ``=` `0`   `    ``# Finding bitwise OR of all elements` `    ``for` `i ``in` `range``(n):` `        ``bits |``=` `arr[i]`   `    ``ans ``=` `bits ``*` `pow``(``2``, n``-``1``)`   `    ``return` `ans`   `# Driver Code` `arr ``=` `[``1``, ``5``, ``6``]` `size ``=` `len``(arr)` `print``(xorSum(arr, size))`   `# This code is contributed by Anant Agarwal.`

## C#

 `// C# approach to finding the XOR_SUM` `using` `System;`   `class` `GFG {` `    `  `    ``// Returns sum of XORs of all subsets` `    ``static` `int` `xorSum(``int` `[]arr, ``int` `n)` `    ``{` `        `  `        ``int` `bits = 0;` `    `  `        ``// Finding bitwise OR of all elements` `        ``for` `(``int` `i = 0; i < n; ++i)` `            ``bits |= arr[i];` `    `  `        ``int` `ans = bits * (``int``)Math.Pow(2, n - 1);` `    `  `        ``return` `ans;` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        `  `        ``int` `[]arr = {1, 5, 6};` `        ``int` `size = arr.Length;` `        `  `        ``Console.Write(xorSum(arr, size));` `    ``}` `}`   `// This code is contributed by Nitin Mittal.`

## PHP

 ``

## Javascript

 ``

Output

`28`

Time complexity: O(n)
Auxiliary space: O(1)

This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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