Why a programmer would prefer O(N^3) instead of O(N^2)

Question

I was studying for my final exam and there is a question in the archive that I cannot find its answer:

The order-of-growth of the running time of one

Answer 1

another thing is, some algorithms have a big constant factor. a O (N^2) might have a big constant factor that won't make it really practical to use ( if N is small enough as kindly noted by Thorban)

Answer 2

Adding to the already posted answers I'd like to mention cache behaviour. A particular memory access pattern might be so much slower due to repeated cache misses that a theoretically slower algorithm with a more cache friendly memory access pattern performs much better.

Answer 3

Here is are examples to convince you that O(N³) can be in some cases better than O(N²).

1. O(N²) algorithm is very complex to code whereas if input size is say N ≤ 100 then for practical use O(N³) can be fast enough

2. O(N²) has a large constant multiplied to it for example c = 1000 hence for N = 100, c⋅N² = 1000⋅100² = 10⁷ whereas if c = 1 for O(N³) then c⋅N³ = 10⁶

3. O(N²) algorithm has very high space complexity as compared to O(N³)

Answer 4

The only reason for choosing one algorithm is not the order-of-growth of the running time. You must analyze:

• How easy it is to implement
• The space required
• How big are the real life inputs (sometimes the time difference is only observable for input sizes that never occurs in reality)

Answer 5

I can think of the following three reasons:

• Ease of initial implementation.
• Ease of maintenance in the future.
• The O(N^3) algorithm may have a lower space complexity than the O(N^2) algorithm (i.e., it uses less memory).

Answer 6

Probably the #1 reason: because the O(N²) algorithm has enough higher constants that for the size of task being contemplated, the O(N³) version is faster.