Yeah, sure! Let's assume the array as bits and you want to find the no of subarrays having a particular bit (say ith bit) set.

so let's assume the above example {3, 8, 13} as,

• for 0th bit, {1, 0, 1}
• for 1th bit, {1, 0, 0}
• for 2nd bit, {0, 0, 1}
• for 3rd bit, {0, 1, 1}

so, now find no of subarrays with odd 1s for each of the bits.

so how can we do this? see we can count no of subarrays starting with 0th index and have odd 1s then we can iterate over all $$$i$$$ letting it to be the starting index of the subarrays.

so let say, for ith index have $$$x$$$ subarrays with odd 1s.

if we encounter 1 at $$$ith$$$ index then for upcoming subarrays the behavior would become opposite (i.e. even 1s subarrays would act like odd 1s subarrays) wo we would do $$${x = n - i - x}$$$ where $$${n - i}$$$ is the number of subarrays starting with $$$ith$$$ index.

So, we will sum-up all the no of subarrays with odd 1s for each bit (say $$$sum_i$$$).

so our answer would be $$${\sum sum_i * 2 ^ i}$$$. here $$$2^i$$$ is the contribution of $$$ith$$$ bit.