Are there any cases where you would prefer a higher big-O time complexity algorithm over the lower one?

Question

Are there are any cases where you would prefer O(log n) time complexity to O(1) time complexity? Or O(n) to O(log n)?

Answer 1

Let's say you're implementing a blacklist on an embedded system, where numbers between 0 and 1,000,000 might be blacklisted. That leaves you two possible options:

1. Use a bitset of 1,000,000 bits
2. Use a sorted array of the blacklisted integers and use a binary search to access them

Access to the bitset will have guaranteed constant access. In terms of time complexity, it is optimal. Both from a theoretical and from a practical point view (it is O(1) with an extremely low constant overhead).

Still, you might want to prefer the second solution. Especially if you expect the number of blacklisted integers to be very small, as it will be more memory efficient.

And even if you do not develop for an embedded system where memory is scarce, I just can increase the arbitrary limit of 1,000,000 to 1,000,000,000,000 and make the same argument. Then the bitset would require about 125G of memory. Having a guaranteed worst-case complexitity of O(1) might not convince your boss to provide you such a powerful server.

Here, I would strongly prefer a binary search (O(log n)) or binary tree (O(log n)) over the O(1) bitset. And probably, a hash table with its worst-case complexity of O(n) will beat all of them in practice.