Approach

To effectively answer the question of how to implement a function to merge two sorted linked lists into a single sorted linked list, we can follow a structured framework:

1. Understand the Problem: Clearly define the task at hand.

2. Plan the Solution: Decide on an algorithmic approach.

3. Write the Code: Implement the solution using a programming language of your choice.

4. Test the Solution: Verify that the implementation works as expected.

5. Optimize: Review the solution for efficiency and clarity.

Key Points

• Clarity of Thought: Ensure you explain your understanding of linked lists and their properties.

• Algorithm Choice: Discuss why you chose a particular algorithm (e.g., iterative vs. recursive).

• Time and Space Complexity: Mention the efficiency of your solution.

• Coding Best Practices: Use clean, well-commented code.

Standard Response

Here’s a sample answer that demonstrates a comprehensive understanding of merging two sorted linked lists:

To merge two sorted linked lists into a single sorted linked list, I would take the following approach:

• Initialization: Create a dummy node that will help in easily managing the merged list.

• Iterate Through Both Lists: Use a pointer to track the current position in the merged list. Compare the values of the current nodes of both lists.

• Select the Smaller Node: Append the smaller node to the merged list and move the pointer in that list to the next node.

• Handle Remaining Nodes: After one of the lists is fully traversed, append the remainder of the other list to the merged list.

• Return the Merged List: Since we used a dummy node, the head of the merged list will be dummy.next.

Here’s the implementation in Python:

class ListNode:
 def __init__(self, val=0, next=None):
 self.val = val
 self.next = next

def mergeTwoLists(l1: ListNode, l2: ListNode) -> ListNode:
 dummy = ListNode(0) # Create a dummy node
 current = dummy # Pointer to construct the new list

 # Traverse both lists
 while l1 and l2:
 if l1.val < l2.val:
 current.next = l1
 l1 = l1.next
 else:
 current.next = l2
 l2 = l2.next
 current = current.next # Move to the next node

 # If there are remaining nodes in either list, append them
 if l1:
 current.next = l1
 if l2:
 current.next = l2

 return dummy.next # Return the merged list starting from the next of dummy

Time Complexity: The time complexity of this algorithm is O(n + m), where n is the number of nodes in the first list and m is the number of nodes in the second list. This is because we are traversing through both lists once.

Space Complexity: The space complexity is O(1), as we are merging the lists in place and not using any additional data structures apart from a few pointers.

Tips & Variations

Common Mistakes to Avoid:

• Not Handling Edge Cases: Failing to consider scenarios where one or both lists are empty.

• Using Excessive Space: Not utilizing the in-place merging could lead to unnecessary memory usage.

• Ignoring Input Validation: Not checking for valid linked list nodes before dereferencing can lead to errors.

Alternative Ways to Answer:

• Recursive Approach: You could also discuss a recursive method to merge the lists, which involves recursively choosing the smaller head and merging the rest.

Role-Specific Variations:

• For Technical Roles: Emphasize the algorithm's efficiency and discuss potential edge cases in detail.

• For Managerial Roles: Focus on the importance of code maintainability and team collaboration when implementing such algorithms.

• For Creative Roles: Highlight how abstract thinking can help in devising unique solutions or optimizations.

Follow-Up Questions:

• What would you do if the linked lists were not sorted?

• **How would