What is the time complexity of pop_front() in a linked list?

• A. O(1)
• B. O(n)
• C. O(log n)
• D. O(n log n)

Answer: (A)

Pop_front removes the first node, and you can update the head pointer to the next node in constant time. Since you don’t need to traverse the list or inspect other nodes, the operation takes the same amount of time regardless of how many elements are stored, so it’s O(1). This is true for singly linked lists as long as you have a pointer to the head (and optionally a way to deallocate the old node). If you had to remove the last element without a back-pointer, you’d need to walk the list to reach it, costing O(n). The O(log n) option doesn’t apply here, as linked-list operations don’t inherently involve logarithmic time unless you’re doing something like a balanced structure or a search.