Get specified bit in addition of two large binary numbers

I am performing an addition operation on two large binary numbers that have an equal number of bits. Both numbers are stored in an array of length $N$, which is rather large.

At first I tried running a loop over them and keeping track of carry bits. This wasted time, because the aim is to get the bit at a specific position in this sum.

So I modified my approach in following way. Starting from the specified index, I am looping until I find a 0 bit in same position on both numbers; I add only those parts to each other and return the bit at the specified position.

This seems okay, but is this the best I can do? Recall that I want to get the value of one specific bit of the sum, given by its position.

Your method will indeed work, but for two $n$-bit numbers it will still take $O(n)$ time. There is a much faster way to solve your problem, with running time $\Theta(\log n)$, known as carry-lookahead addition. The downside is that in my experience teaching this technique, people have a hard time wrapping their heads around it. You can find online sources, here and here, and some algorithm or computer organization texts cover it. It's a really interesting algorithm, well worth taking the time to figure it out.

StackExchange Computer Science Q#4643, answer score: 3