Suppose a circular queue of capacity (n – 1) elements is implemented with an array of n elements. Assume that the insertion and deletion operation are carried out using REAR and FRONT as array index variables, respectively. Initially, REAR = FRONT = 0. The conditions to detect queue full and queue empty are**(A)** Full: (REAR+1) mod n == FRONT, empty: REAR == FRONT**(B)** Full: (REAR+1) mod n == FRONT, empty: (FRONT+1) mod n == REAR**(C)** Full: REAR == FRONT, empty: (REAR+1) mod n == FRONT**(D)** Full: (FRONT+1) mod n == REAR, empty: REAR == FRONT**Answer:** **(A)****Explanation:**

Suppose we start filling the queue. Let the maxQueueSize ( Capacity of the Queue) is 4. So the size of the array which is used to implement this circular queue is 5, which is n. In the beginning when the queue is empty, FRONT and REAR point to 0 index in the array. REAR represents insertion at the REAR index. FRONT represents deletion from the FRONT index. enqueue("a"); REAR = (REAR+1)%5; ( FRONT = 0, REAR = 1) enqueue("b"); REAR = (REAR+1)%5; ( FRONT = 0, REAR = 2) enqueue("c"); REAR = (REAR+1)%5; ( FRONT = 0, REAR = 3) enqueue("d"); REAR = (REAR+1)%5; ( FRONT = 0, REAR = 4) Now the queue size is 4 which is equal to the maxQueueSize. Hence overflow condition is reached. Now, we can check for the conditions. When Queue Full : ( REAR+1)%n = (4+1)%5 = 0 FRONT is also 0. Hence ( REAR + 1 ) %n is equal to FRONT. When Queue Empty : REAR was equal to FRONT when empty ( because in the starting before filling the queue FRONT = REAR = 0 ) Hence Option A is correct.

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