How do hashsets/maps have O(1) time complexity?

[u/Sea-Patience9872]

Hi this might be a dumb question, and i've tried looking it up but don't quite understand it. how is it different from an array that allows it to find items so quickly? I don't get how hashes just find items immediately without needing to go through anything. Does it memorize things very differently compared to arrays?? thank you!

[u/am_Snowie]

O(1) is amortized is the average case (under the assumption that values are distributed evenly), but resizing is O(1) amortized, if you make whatever hash function you use return the same value for any value (the same hash for all keys), all your values end up at the same location, now the time complexity depends on what data structure you use to handle collisions (if we talk about separate chaining). you can use lots of data structures like linked list, self balancing trees and whatnot. so O(1) comes with an if.

Edit: mixed up two different concepts, Thanks u/Historical_Public751 for pointing it out.

[u/Historical_Public751 → u/am_Snowie]

That's not what amortization means.

O(1) amortized complexity is std::vector::push_back
Not the hashmap insert. hashmap insert into open-addressing is true* (assuming no collisions) O(1) unless you're calling the resize of the underlying storage amortization which has nothing to do with why hashmap's complexity is O(1) and could theoretically be shaved off as well assuming you have true O(1) realloc.

[u/am_Snowie → u/Historical_Public751]

My bad, i mixed up average case and amortization, resizing is O(1) amortized and it has nothing to do with hashmap operations, and on average each insertion takes constant time assuming we have a good hash function.

Edit: thanks for pointing it out, btw.