Question

Consider the following sequence of operations on an empty stack and an empty queue.
Stack:
push(54); push(52); pop(); push(55); push(62); s = pop();
Queue:
enqueue(21); enqueue(24); dequeue(); enqueue(28); enqueue(32); q = dequeue();
The value of \( s + q \) is \(\underline{\hspace{2cm}}\).

Hint:

Stacks follow LIFO order, while queues follow FIFO order.

Answer 1

Correct Answer: 86

Step 1: Perform stack operations.

push(54) → [54]
push(52) → [54, 52]
pop() → removes 52
push(55) → [54, 55]
push(62) → [54, 55, 62]
pop() → removes 62

Therefore, s = 62

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Step 2: Perform queue operations.

enqueue(21) → [21]
enqueue(24) → [21, 24]
dequeue() → removes 21
enqueue(28) → [24, 28]
enqueue(32) → [24, 28, 32]
dequeue() → removes 24

Therefore, q = 24

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Step 3: Compute the final result.

s + q = 62 + 24 = 86

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Final Answer:

86