In 620C - Pearls in a Row I watch some source code of other people. Most of them used map. They iterate for n times and count the element of the array using map when they find that the value of current element in the map is 2 then they clear the whole map.. For that I am bit confused about that solution. It will be pleasure if anyone tell me what is the actual complexity of that solution :)
to delete all the elemants in map take O(N) where N is size of map
sum of all the sizes of map during program, about what you have said, is N
so such solution as you say take O(N + NlogN)
O(N + NlogN) = O(NlogN)
Best way to do it is using set just because we dont need to count elements : we only need to know if this element was before. Anyway, in java you can clear map(in my case set) with O(1) complexity like this:
BTW, if we change it to HashSet(unordered_set in c++), we can get O(n) complexity.
If I'm not mistaken, old set will be cleared away by GC, and it will still take O(n) time
Old set will be cleared by GC of corse but please explain me why do you think that it take O(n) time?
I don't know right answer, but I can share my thoughts: complexity is O(1) because we don't remove each element separately, we remove the whole object from the memory.
Anyway, I made an experiment and get some data which confirms my thesis: