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

There can be many reasons to prefer an algorithm with higher big O time complexity over the lower one:

• most of the time, lower big-O complexity is harder to achieve and requires skilled implementation, a lot of knowledge and a lot of testing.
• big-O hides the details about a constant: algorithm that performs in 10^5 is better from big-O point of view than 1/10^5 * log(n) (O(1) vs O(log(n)), but for most reasonable n the first one will perform better. For example the best complexity for matrix multiplication is O(n^2.373) but the constant is so high that no (to my knowledge) computational libraries use it.
• big-O makes sense when you calculate over something big. If you need to sort array of three numbers, it matters really little whether you use O(n*log(n)) or O(n^2) algorithm.
• sometimes the advantage of the lowercase time complexity can be really negligible. For example there is a data structure tango tree which gives a O(log log N) time complexity to find an item, but there is also a binary tree which finds the same in O(log n). Even for huge numbers of n = 10^20 the difference is negligible.
• time complexity is not everything. Imagine an algorithm that runs in O(n^2) and requires O(n^2) memory. It might be preferable over O(n^3) time and O(1) space when the n is not really big. The problem is that you can wait for a long time, but highly doubt you can find a RAM big enough to use it with your algorithm
• parallelization is a good feature in our distributed world. There are algorithms that are easily parallelizable, and there are some that do not parallelize at all. Sometimes it makes sense to run an algorithm on 1000 commodity machines with a higher complexity than using one machine with a slightly better complexity.
• in some places (security) a complexity can be a requirement. No one wants to have a hash algorithm that can hash blazingly fast (because then other people can bruteforce you way faster)
• although this is not related to switch of complexity, but some of the security functions should be written in a manner to prevent timing attack. They mostly stay in the same complexity class, but are modified in a way that it always takes worse case to do something. One example is comparing that strings are equal. In most applications it makes sense to break fast if the first bytes are different, but in security you will still wait for the very end to tell the bad news.
• somebody patented the lower-complexity algorithm and it is more economical for a company to use higher complexity than to pay money.
• some algorithms adapt well to particular situations. Insertion sort, for example, has an average time-complexity of O(n^2), worse than quicksort or mergesort, but as an online algorithm it can efficiently sort a list of values as they are received (as user input) where most other algorithms can only efficiently operate on a complete list of values.

Answer 2

Consider a red-black tree. It has access, search, insert, and delete of O(log n). Compare to an array, which has access of O(1) and the rest of the operations are O(n).

So given an application where we insert, delete, or search more often than we access and a choice between only these two structures, we would prefer the red-black tree. In this case, you might say we prefer the red-black tree's more cumbersome O(log n) access time.

Why? Because the access is not our overriding concern. We are making a trade off: the performance of our application is more heavily influenced by factors other than this one. We allow this particular algorithm to suffer performance because we make large gains by optimizing other algorithms.

So the answer to your question is simply this: when the algorithm's growth rate isn't what we want to optimize, when we want to optimize something else. All of the other answers are special cases of this. Sometimes we optimize the run time of other operations. Sometimes we optimize for memory. Sometimes we optimize for security. Sometimes we optimize maintainability. Sometimes we optimize for development time. Even the overriding constant being low enough to matter is optimizing for run time when you know the growth rate of the algorithm isn't the greatest impact on run time. (If your data set was outside this range, you would optimize for the growth rate of the algorithm because it would eventually dominate the constant.) Everything has a cost, and in many cases, we trade the cost of a higher growth rate for the algorithm to optimize something else.

Answer 3

People have already answered your exact question, so I'll tackle a slightly different question that people may actually be thinking of when coming here.

A lot of the "O(1) time" algorithms and data structures actually only take expected O(1) time, meaning that their average running time is O(1), possibly only under certain assumptions.

Common examples: hashtables, expansion of "array lists" (a.k.a. dynamically sized arrays/vectors).

In such scenarios, you may prefer to use data structures or algorithms whose time is guaranteed to be absolutely bounded logarithmically, even though they may perform worse on average.
An example might therefore be a balanced binary search tree, whose running time is worse on average but better in the worst case.

Answer 4

There is always the hidden constant, which can be lower on the O(log n) algorithm. So it can work faster in practice for real-life data.

There are also space concerns (e.g. running on a toaster).

There's also developer time concern - O(log n) may be 1000× easier to implement and verify.

Answer 5

I'm surprised nobody has mentioned memory-bound applications yet.

There may be an algorithm that has less floating point operations either due to its complexity (i.e. O(1) < O(log n)) or because the constant in front of the complexity is smaller (i.e. 2n² < 6n²). Regardless, you might still prefer the algorithm with more FLOP if the lower FLOP algorithm is more memory-bound.

What I mean by "memory-bound" is that you are often accessing data that is constantly out-of-cache. In order to fetch this data, you have to pull the memory from your actually memory space into your cache before you can perform your operation on it. This fetching step is often quite slow - much slower than your operation itself.

Therefore, if your algorithm requires more operations (yet these operations are performed on data that is already in cache [and therefore no fetching required]), it will still out-perform your algorithm with fewer operations (which must be performed on out-of-cache data [and therefore require a fetch]) in terms of actual wall-time.

Answer 6

Alistra nailed it but failed to provide any examples so I will.

You have a list of 10,000 UPC codes for what your store sells. 10 digit UPC, integer for price (price in pennies) and 30 characters of description for the receipt.

O(log N) approach: You have a sorted list. 44 bytes if ASCII, 84 if Unicode. Alternately, treat the UPC as an int64 and you get 42 & 72 bytes. 10,000 records--in the highest case you're looking at a bit under a megabyte of storage.

O(1) approach: Don't store the UPC, instead you use it as an entry into the array. In the lowest case you're looking at almost a third of a terabyte of storage.

Which approach you use depends on your hardware. On most any reasonable modern configuration you're going to use the log N approach. I can picture the second approach being the right answer if for some reason you're running in an environment where RAM is critically short but you have plenty of mass storage. A third of a terabyte on a disk is no big deal, getting your data in one probe of the disk is worth something. The simple binary approach takes 13 on average. (Note, however, that by clustering your keys you can get this down to a guaranteed 3 reads and in practice you would cache the first one.)