Question

We want to build a queue for integer elements that in addition to the operations we saw during the lecture, also supports a seeSaw() operation that on a given queue holding the elements Q0,Q1,…, Qn−1 returns Q0+1/Q1+Q2+1/Q3+⋯ All operations should run in O(1) time. Your data structure should take O(n) space, where n is the number of elements currently stored in the data please explain and write psuedocode if I enter these the program should give this: enqueue(1) enqueue(2) enqueue(3) seeSaw() enqueue(4) dequeue() enqueue(5) enqueue(6) seeSaw() dequeue() seeSaw() will return 1+1/2+3=4.5,2+1/3+4+1/5+6=12.5333,3+1/4+5+1/6=8.41666 The time complexity including seeSaw() must be O(1) !!!

Answer 1

To build a queue with enqueue, dequeue, and seeSaw operations in O(1) time, we can use a combination of two stacks. The pseudocode provided demonstrates how to implement this data structure.

Step-by-step explanation:

To build a queue that supports enqueue, dequeue, and seeSaw operations with O(1) time complexity, we can use a combination of two stacks. We'll call these stacks S1 and S2. When enqueueing an element, we simply push it onto stack S1. When dequeueing an element, we check if S2 is empty. If it is, we pop all elements from S1 and push them onto S2. Finally, we pop the top element from S2. The seeSaw operation can be performed by summing up the elements of S2 and dividing by the size of S1. We can implement this data structure using the following pseudocode:

enqueue(element):
S1.push(element)

dequeue():
if S2 is empty:
while S1 is not empty:
S2.push(S1.pop())
return S2.pop()

seeSaw():
sum = 0
for each element in S2:
sum += element
return sum / S1.size()