GATE CS 2013 Q44 — Arrays, Stacks and Queues

To find the worst-case time complexity of a sequence of $n$ queue operations starting with an empty queue, we can use amortized analysis (specifically, the aggregate method).

Let's analyze the cost of each operation in the sequence:

1. Enqueue(Q): Takes $O(1)$ time and adds exactly one element to the queue.
2. Dequeue(Q): Takes $O(1)$ time and removes exactly one element from the queue.
3. MultiDequeue(Q): Removes up to $k$ elements. The time complexity is proportional to the number of elements actually dequeued, say $r$ (where $r \le k$), plus a constant overhead $O(1)$ for initialization and checks. Thus, its cost is $O(r + 1)$.

Key Observation:

An element can only be dequeued if it has been previously enqueued.

• In any sequence of $n$ operations, the maximum number of Enqueue operations is at most $n$.
• Therefore, at most $n$ elements can ever be enqueued.
• Consequently, the total number of elements dequeued across all Dequeue and MultiDequeue operations in the entire sequence cannot exceed $n$.

Total Cost Calculation:

The total time complexity for the sequence of $n$ operations is the sum of:

• The constant overhead for each of the $n$ operations: $n \times O(1) = O(n)$.
• The actual cost of all enqueues and dequeues: Since there are at most $n$ enqueues and at most $n$ dequeues, this total cost is bounded by $O(n)$.

Thus, the total worst-case time complexity for $n$ operations is $O(n)$. Since executing $n$ operations must take at least $\Omega(n)$ time, the tight bound is $\Theta(n)$.