Consider an array-based queue on which you need to perform a dequeue…
2024
Consider an array-based queue on which you need to perform a dequeue operation. You can implement this either using stack operations (push and pop) or using the queue's own native operations (enqueue and dequeue) directly. Assume both approaches always return the same output. Identify the correct difference between the two approaches.
- A.
They will have different time complexities
- B.
The memory used will not be different
- C.
There are chances that output might be different
- D.
No differences
Show answer & explanation
Correct answer: A
A queue guarantees FIFO removal. Two ways to realize a queue's dequeue return the SAME sequence of elements but cost different amounts of work per call. With the two-stack simulation (an input stack and an output stack), an element is pushed once onto the input stack and, whenever the output stack runs dry, transferred -- popped from the input stack and pushed onto the output stack -- before it can finally be popped and returned. A properly-implemented native queue (tracked via a front index/pointer, not a naive shifting array) instead removes the front element directly in one step on every single call, with no transfer step at all.
Enqueue a, b, c using only push: the input stack holds [a, b, c] with c on top.
First dequeue: the output stack is empty, so every element is transferred — pop c, push c; pop b, push b; pop a, push a — onto the output stack, which now holds [c, b, a] with a on top; pop a and return it. This single dequeue did O(n) work (n = 3 transfers).
Second dequeue: the output stack already holds [c, b] with b on top, so simply pop and return b — O(1) work, no transfer needed.
A native queue with a front index instead removes a in one O(1) step for every dequeue call, including the very first one, since no reversal is ever required.
Averaged over many operations, the stack-based transfers amortize to O(1) per dequeue, since each element is pushed and popped at most twice across its lifetime -- but any SINGLE dequeue call that finds the output stack empty costs O(n) in the worst case, a spike a properly-implemented native queue's dequeue never has (it is always exactly O(1), call after call). So it is specifically the worst-case, per-call time complexity that differs between the two approaches, even though -- as the question guarantees -- both always return identical output and both cost O(1) on amortized average. The stack-based approach also needs a second auxiliary stack that the native queue never allocates.
The genuine, correct-option-matching distinction between the two approaches is that their worst-case time complexities differ -- that is what "different time complexities" refers to here.