3.4. queue via stack
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# 3.4. Queue via stacks
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> Implement a MyQueue class which implements a queue using two stacks
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## First idea
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A stack is LIFO, first element to get out is the one inserted last. And queue is FIFO.
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## Hints
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* How to remove the oldest item from a stack if you only have access to the newest item?
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* Delete the one opposite in the array, so the last one...
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* The trick must be inverting something because I need two stacks...
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* We can remove the oldest elem in a stack by repeatedly removing the newest item (inserting those in the temporary stack, stack \#2) until we get down to one element, then put all the elems back. This requires O(n). Can we optimize for scenarios where we might do several pops in a row?
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* If we could check if an elem is present in stack \#2 yes, but there's no such function
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## Solution
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We need to modify `peek` and `pop` to work in reverse order. We can use stack \#2 to reverse the order of the elements by popping s1 and pushing the elements on to s2. In each `peek` and `pop` operation, we pop everything from s1 onto s2, perform `peek` and `pop` and then push everything back.
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If two `peek`/`pop` must be performed, we need to move everything again. In a lazy approach, all elemenst remain in s2 until they must absolutely be reversed.
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stack_newest has the newest elements on top and stack_oldest has the oldest elements on top. When we dequeue an element, we dequeue from stack_oldest. If empty, transfer all elemens from stack_newest into stack_oldest in reverse order.
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[Code in Github](https://github.com/careercup/CtCI-6th-Edition/blob/master/Java/Ch%2003.%20Stacks%20and%20Queues/Q3_04_Queue_via_Stacks/MyQueue.java)
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