Here’s the revised version with corrected grammar:

Hi all, hope everything is going well. I tried to solve the problem “Sum of XOR of all possible subsets” on LeetCode and found a more efficient approach than generating all subsets, as explained on GeeksForGeeks. However, I’m struggling to understand the mathematical aspect of it. I even tried using GPT, but it still didn’t clarify things completely.

An efficient approach involves identifying patterns with respect to XOR properties. Consider subsets represented in binary form:

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

By analyzing the binary representations of these XORs, we observe that each bit position from 0 to n-1 contributes exactly half of 2^n to the total sum. We can leverage this observation to impose two conditions for each bit position i:

1. If any element in arr[] has the ith bit set, exactly half of 2^n subsets will contribute to the sum, totaling 2^(n-1+i).
2. If no element in arr[] has the ith bit set, that bit will not contribute to any subset's sum.

The proof is as follows:

Case 1:
Let's assume there are k elements in the array with the ith bit set (where k > 0). To have a subset with the ith bit set in its XOR, we need an odd number of elements with the 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 the sum becomes 2^(n+i-1).

Case 2:
If no element has the ith bit set (i.e., k = 0), the contribution of the ith bit towards the total sum remains 0.

Now, instead of iterating through all elements one by one, we use a trick: simply take the bitwise OR of all values in arr[] and multiply by 2^(n-1).

I now understand the bitwise OR part.

Bitwise OR sets each bit to 1 if any number in the array has a 1 in that bit position.

1 in binary: 001
5 in binary: 101
6 in binary: 110
Performing OR on these:

001 | 101 = 101
101 | 110 = 111

The OR result is 111 (which is 7 in decimal).

What’s next? Essentially, how do we determine that multiplying 7 by 2^(n-1) is the next step? Why?

I’ve refined the text for clarity and corrected any grammar issues. Let me know if there’s anything else you’d like to adjust!