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Mediumqueuecircular

~90s

Elements in a circular queue

A circular queue occupies an array of 8 slots, with one slot always left empty so that full and empty can be told apart. front indexes the first element and rear indexes the next free slot. If front = 4 and rear = 3, how many elements does the queue hold?

Show the answer and the worked solution

Correct answer: C.

7

Worked solution

  1. With rear pointing at the next free slot, the number of elements is (rear − front + n) mod n.
  2. Substituting the values gives (3 − 4 + 8) mod 8 = 7 mod 8 = 7.
  3. Since one slot is deliberately left empty, 7 is the largest an eight-slot array can hold, so the queue is in fact full.

Answer: 7.

The idea

Adding n before taking the modulus keeps the arithmetic correct when rear has wrapped past front. Leaving one slot unused is the standard trick that makes (rear + 1) mod n == front mean 'full'.

Shortcut. rear sits one slot behind front, (3 + 1) mod 8 = 4, which is the definition of full in the one-empty-slot scheme: capacity 8 − 1 = 7.

Common trap. 1 takes the gap front − rear = 4 − 3 without adding n, measuring the empty stretch instead of the occupied one.

Data Structures & Algorithms

6 of 40

Level

Medium

Target time

~90s

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Elements in a circular queue — Aptitude Question

Elements in a circular queue: A circular queue occupies an array of 8 slots, with one slot always left empty so… Answer and worked solution on CodeKairo.

Question

A circular queue occupies an array of 8 slots, with one slot always left empty so that full and empty can be told apart. front indexes the first element and rear indexes the next free slot. If front = 4 and rear = 3, how many elements does the queue hold?

Options

  1. 0
  2. 1
  3. 7
  4. 8
Show the answer and the worked solution

Correct answer: C. 7

Worked solution

  1. With rear pointing at the next free slot, the number of elements is (rear − front + n) mod n.
  2. Substituting the values gives (3 − 4 + 8) mod 8 = 7 mod 8 = 7.
  3. Since one slot is deliberately left empty, 7 is the largest an eight-slot array can hold, so the queue is in fact full.

Answer: 7.

The idea

Adding n before taking the modulus keeps the arithmetic correct when rear has wrapped past front. Leaving one slot unused is the standard trick that makes (rear + 1) mod n == front mean 'full'.

Shortcut. rear sits one slot behind front, (3 + 1) mod 8 = 4, which is the definition of full in the one-empty-slot scheme: capacity 8 − 1 = 7.

Common trap. 1 takes the gap front − rear = 4 − 3 without adding n, measuring the empty stretch instead of the occupied one.

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