[csnweb]

If you replace an n^2 algorithm with a log(n) lookup you get an exponential speed up. Although a hashmap lookup is usually O(1), which is even faster.

[ryao]

That is not true unless n^C / e^n = log(n) where C is some constant, which it is not. The difference between log(n) and some polynomial is logarithmic, not exponential.

[csnweb]

But if you happen to have n=2^c, then an algorithm with logarithmic complexity only needs c time. Thats why this is usually referred to as exponential speedup in complexity theory, just like from O(2^n) to O(n). More concretely if the first algorithm needs 1024 seconds, the second one will need only 10 seconds in both cases, so I think it makes sense.

[wasabi991011]

It depends if you consider "speedup" to mean dividing the runtime or applying a function to the runtime.

I.e. you are saying and f(n) speedup means T(n)/f(n), but others would say it means f(T(n)) or some variation of that.

[ndriscoll]

They're still using the map in a loop, so it'd be nlogn for a tree-based map or n for a hash map.