Computers and Technology

Computers and Technology, 24.06.2019 00:20, talyku843

Describe a data structures that supports the stack push and pop operations and a third operation findmin, which returns the smallest element in the data structure, all in o(1) worst-case time.

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Computers and Technology, 27.09.2019 22:30, danielcano12281621

Given an initially empty stack s that accepts integers, the following operations are performed: s.push (10) s.push (20) s.push (30) s.push (40) s.pop () s.pop () s.push (50) s.push (60) s.push (70) s.pop () s.pop () s.push (80) s.push (90) s.pop () a) write out the composition of the stack after these operations. b) what sequences of integers was popped off the stack?

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Computers and Technology, 28.09.2019 00:00, taylorannsalazar

In python - create a stack class that uses a list to store the stack's data. it needs to include these methods: push(integer) pop() - which returns the element at the top of the stack checksize() - which returns the count of elements in the stack

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Computers and Technology, 09.10.2019 20:00, spookymod4845

Say that we want to maintain both a stack and a queue of the same elements. a new element that arrives implies a push to the stack and a enqueue to the queue. any pop operation should remove the element from the queue and from the stack as well (even though its not the highest propriety element). dequeue removes the head of the queue, but also removes this element from the stack (even though its not the last to being inserted). using two copies of every element and using additional pointers between element, show that we can implement both a stack and a queue so that every operation requires o(1) time.

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Computers and Technology, 16.11.2019 01:31, weberalycia

Give a python implementation for the midstack adt. the midstack adt supports the following operations * midstack (): initializes an empty midstack object . mids.is_empty (): returns true if s does not contain any elements, or false otherwise len (mids): returns the number of elements mids mids.push (e) : adds element e to the top of mids. mids.top (): returns a reference to the top element ofmids, without removing it; an exception is raised if s is empty mids.pop ): removes and returns the top element from mids; an exception is raised ifmids is empty . * e mids.mid push (e) : adds element e in the middle of mids that is, assuming there are n elements in s: in the case n is even, e would go exactly in the middle. if n is odd, e will go after the "2th element. for example, your implementation should follow the behavior as demonstrated in the two execution examples below: > > > míds = midstack() > > > mids.push (2) > > > mids.push (4) > > > mids.push (6) > > > mids.push (8) > > > mids.mid push (10) > > > mids.pop () > > > midsmidstack ) > > > mids.push (2) > > > mids.push (4) > > > mids.push (6) > > > mids.push (8) > > > mids.push (10) > > > mids.mid push (12) > > > mids.pop () 10 > > > mids.pop () > > > mids.pop () 6 > > > mids.pop () 10 > > > mids.pop () > > > mids.pop ) 12 > > > mids.pop () > > > mids.pop () 2 > > > mids.pop( > > > mids.pop () 2

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