A. It removes the last element instead of the first when dequeuing.
B. It appends elements incorrectly to the front.
C. It accesses an invalid index causing runtime error.
D. It does not print the queue correctly.
Solution
Step 1: Understand the dequeue operation
Dequeue should remove the first element (front) from the queue.
Step 2: Analyze the slicing used
queue = queue[:len(queue)-1] removes the last element, not the first.
Final Answer:
It removes the last element instead of the first when dequeuing. -> Option A
Quick Check:
Dequeue removes front, not last [OK]
Hint: Dequeue removes front: slice from index 1, not slice end [OK]
Common Mistakes:
Removing last element instead of first
Confusing append with dequeue operation
Not updating queue slice after removal
5. You want to process tasks in the order they arrive but also need to handle urgent tasks immediately. Which data structure combination best solves this problem?
hard
A. Use a single queue and insert urgent tasks at the front manually.
B. Use a list and sort tasks by urgency before processing.
C. Use a stack to process urgent tasks and a queue for normal tasks, alternating between them.
D. Use two queues: one for urgent tasks and one for normal tasks, processing urgent queue first.
Solution
Step 1: Understand the problem requirements
Tasks must be processed in arrival order, but urgent tasks get priority.
Step 2: Evaluate data structure options
Two queues separate urgent and normal tasks; urgent queue is processed first, preserving order.
Step 3: Why other options fail
Inserting at front breaks FIFO; stack reverses order; sorting loses arrival order.
Final Answer:
Use two queues: one for urgent tasks and one for normal tasks, processing urgent queue first. -> Option D
Quick Check:
Two queues handle priority and order [OK]
Hint: Separate urgent and normal tasks in two queues [OK]
Common Mistakes:
Trying to insert urgent tasks at front of one queue
Using stack which reverses order
Sorting tasks losing original arrival order
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