GATE CS 2012 Q.35 (CS Q.35): Stacks and Queues

Suppose a circular queue of capacity ( $n - 1$ ) elements is implemented with an array of $n$ elements. Assume that the insertion and deletion operations are carried out using REAR and FRONT as array index variables, respectively. Initially, $\text{REAR} = \text{FRONT} = 0$ . The conditions to detect queue full and queue empty are

full: $(\text{REAR} + 1) \bmod n == \text{FRONT}$
empty: $\text{REAR} == \text{FRONT}$

full: $(\text{REAR} + 1) \bmod n == \text{FRONT}$
empty: $(\text{FRONT} + 1) \bmod n == \text{REAR}$

full: $\text{REAR} == \text{FRONT}$
empty: $(\text{REAR} + 1) \bmod n == \text{FRONT}$

full: $(\text{FRONT} + 1) \bmod n == \text{REAR}$
empty: $\text{REAR} == \text{FRONT}$

Official answer and solution

Official answer: A
From: Official GATE 2012 answer key

Solution

Evaluate queue full and empty conditions in an array-based circular queue with one slot left unused.

In an array of size $n$ , capacity is restricted to $n - 1$ elements to unambiguously distinguish between queue empty and queue full conditions.
Initially, both indices point to the same index:

\text{Empty condition: } \text{REAR} == \text{FRONT}

As elements are enqueued, REAR is incremented circularly using modular arithmetic: $(\text{REAR} + 1) \bmod n$ .
When $n - 1$ elements are stored in the queue, advancing REAR by one more step would cause it to reach FRONT. Thus, the queue full condition is detected when advancing REAR matches FRONT:

\text{Full condition: } (\text{REAR} + 1) \bmod n == \text{FRONT}

This confirms option (A).

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2012 Computer Science and Information Technology (CS)
Question: 35 of 65, printed as CS Q.35
Section: Computer Science and Information Technology

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