@ Juver,

I got the following response in the group digest mail. It'll be great
if you can provide some inputs so I can refine my understanding...

Could you please help me understand how my delete operation takes O(n)
time for every delete? I propose to maintain the minimum at the queue
level (one variable for the entire queue object) which can change on
each insert (O(1) operation to see if the newly inserted item is the
new minimum) and on delete of a current min item, a O(n) list
traversal is required to figure out what the min is now... if the
deleted item is not the current min, no updates to current minimum
value need to be made...

I did a little bit of reading on amortized complexity. You're
definitely correct. I don't know as much as you probably do on that
subject.
But is the following statement about the 2 stack implementation
correct? After inserting a bunch of items, say 10000, in your queue
(which fills up stack 1), lets say you do delete_rear (u need to pop
all items from stack 1 and push them in stack 2) and then a
delete_front (pop from second stack and push all into first) in
succession until the queue is empty, wont each operation cost O(n)?

Based on my preliminary understanding, to estimate amortized
complexity, we can ignore O(n) operations provided they occur only
once in every K operations... but when there's a chance that they can
occur for every 1 operation, that's the worst case scenario there.
would it still be amortized...

Pls Note: this is just to better my understanding.. am not challenging
anybody...


---------------------------------------------------------------------------------------
juver++ <avpostni...@gmail.com> Jan 10 06:25AM -0800 ^

You should analyze your algo more precisely and study something about
amortized time complexity.
Your delete operation takes O(n) time for EVERY query.
So for the sequence of M deletetions there is an average time O(N)
which is
NOT constant on the worst case.

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