A stack is implemented using queue(s); q1 is the queue that stores the stack's…
2023
A stack is implemented using queue(s); q1 is the queue that stores the stack's elements. The Java method below is one of the operations of that stack.
public void fun(int x) {
q1.offer(x);
}
Which stack operation does this method perform, and in this implementation which operation - push() or pop() - is the costlier one?
Answer: B. Perform push() with pop as the costlier operation — ConceptA stack is governed by LIFO: the most recently inserted element must be removed first. A queue is governed by FIFO: offer(x) appends x at the rear,…
- A.
Perform push() with push as the costlier operation
- B.
Perform push() with pop as the costlier operation
- C.
Perform pop() with push as the costlier operation
- D.
More than one of the above
- E.
None of the above
Attempted by 730 students.
Show answer & explanation
Correct answer: B
Concept
A stack is governed by LIFO: the most recently inserted element must be removed first. A queue is governed by FIFO: offer(x) appends x at the rear, while removal normally occurs at the front.
When a stack is built from ordinary queue operations, the LIFO reordering work must be paid either during insertion or during removal. A direct enqueue makes insertion constant-time and leaves that reordering work for removal.
Application
The statement q1.offer(x) appends x to q1 and increases the number of stored elements by one; it therefore implements the stack insertion operation push(x).
Starting from an empty queue, push(10), push(20), and push(30) produce q1 = [10, 20, 30]. Each call performs one offer operation, so push() takes O(1) time.
A stack pop must return 30, but the queue front is 10. The implementation must move 10 and 20 out of the way (or rotate them) before removing 30.
Moving up to n − 1 earlier elements makes pop() take O(n) time in this costly-pop design.
Cross-check
Queue-based design | Insertion work | Removal work |
|---|---|---|
Costly push | Reorders existing elements during push(), O(n) | Removes the prepared front element, O(1) |
Costly pop | Directly enqueues with offer(), O(1) | Moves earlier elements to reach the newest one, O(n) |
“Perform push() with push as the costlier operation” would require fun(x) itself to rotate or re-enqueue existing elements; the shown body does not do that.
“Perform pop() with push as the costlier operation” conflicts with the state change because offer(x) inserts x rather than removing an element.
“More than one of the above” would require the same fixed method to be both insertion and removal, or to assign two incompatible cost profiles at once.
“None of the above” would require every explicit pairing to be incompatible with the method, but a direct-enqueue insertion pairing is offered.
Result
The method performs push(), and pop() is the costlier operation: push() is O(1), whereas pop() is O(n).