Solution Approach

The solution uses a two-pointer simulation approach to perform the merge operation. Here's the step-by-step implementation:

Step 1: Initialize two pointers

p = q = list1

Both pointers p and q start at the head of list1. These will help us locate the critical connection points.

Step 2: Position pointer p before the removal range

for _ in range(a - 1):  p = p.next

Move p forward (a-1) times. After this loop, p points to the node just before the a-th node. This is crucial because we need to modify p.next to point to list2.

Step 3: Position pointer q at the end of the removal range

for _ in range(b):  q = q.next

Move q forward b times from the head. After this loop, q points to the b-th node (the last node to be removed).

Step 4: Connect the pre-removal node to list2

p.next = list2

This disconnects the nodes from position a to b and starts the connection to list2.

Step 5: Find the tail of list2

while p.next:  p = p.next

Since p.next now points to list2, we traverse through list2 until we reach its tail node (where p.next becomes None).

Step 6: Complete the connection

p.next = q.next q.next = None

Connect the tail of list2 to the node that came after the b-th node in the original list1. Setting q.next = None helps clean up the removed segment (though this is optional since those nodes are no longer reachable).

Step 7: Return the modified list

return list1

The head of list1 remains unchanged, so we return it as the result.

Time Complexity: O(a + b + m) where m is the length of list2. We traverse to position a-1, then to position b, and finally through all nodes of list2.

Space Complexity: O(1) as we only use a constant amount of extra space for the pointers.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Given:

• list1: 10 → 20 → 30 → 40 → 50
• list2: 100 → 200
• a = 2, b = 3 (remove nodes at positions 2 and 3)

Goal: Remove nodes 20 and 30 from list1 and insert list2 in their place.

Step 1: Initialize pointers

p = q = list1 Both p and q point to node 10

Step 2: Position p before removal range (move a-1 = 1 time)

p moves 1 step: p now points to node 10 list1: 10 → 20 → 30 → 40 → 50  p

Step 3: Position q at end of removal range (move b = 3 times)

q moves 3 steps: 10 → 20 → 30 list1: 10 → 20 → 30 → 40 → 50  p q

Step 4: Connect p.next to list2

p.next = list2 (node 100) Now: 10 → 100 → 200  p (Nodes 20 and 30 are disconnected)

Step 5: Find tail of list2

Move p through list2: p: 10 → 100 → 200  p p (final position)

Step 6: Complete the connection

p.next = q.next (node 40) Final: 10 → 100 → 200 → 40 → 50

Result: The modified list is 10 → 100 → 200 → 40 → 50

The nodes at positions 2 and 3 (values 20 and 30) have been successfully replaced with the entire list2 (100 → 200).

Time and Space Complexity

Time Complexity: O(m + n), where m is the length of list1 and n is the length of list2.

The time complexity breaks down as follows:

• First loop: O(a) iterations to position pointer p at index a-1
• Second loop: O(b) iterations to position pointer q at index b
• Third loop: O(n) iterations to traverse through all nodes of list2 to find its tail
• Since a and b are at most m-1, the worst case for the first two loops is O(m)
• Total: O(m) + O(n) = O(m + n)

Space Complexity: O(1)

The algorithm only uses a constant amount of extra space:

• Two pointers p and q that reference existing nodes
• No additional data structures are created
• The operation modifies the existing linked lists in-place by adjusting pointers

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Off-by-One Error in Pointer Positioning

The Pitfall: The most common mistake is incorrectly positioning the before_removal_node pointer. Many developers mistakenly move it a times instead of a-1 times, which would position it at the a-th node rather than the node before it.

Incorrect Code:

# WRONG: This moves the pointer TO position a, not BEFORE it for _ in range(a): # Should be (a-1)  before_removal_node = before_removal_node.next

Why This Fails:

• If you move a times, before_removal_node points to the first node that should be removed
• When you then set before_removal_node.next = list2, you're keeping one node that should have been removed
• This results in an extra node in the final list

The Solution: Always remember that to insert at position a, you need to access the node at position a-1. Use range(a-1) for the first pointer.

2. Edge Case: When a = 1 (Removing from the Beginning)

The Pitfall: When a = 1, the code tries to move the pointer (a-1) = 0 times, which means before_removal_node stays at the head. This works correctly for the connection but conceptually might be confusing.

Potential Issue: If someone modifies the code without understanding this edge case, they might add special handling that breaks the solution:

# WRONG: Unnecessary special case handling if a == 1:  # Trying to handle first node removal differently  temp = list1  for _ in range(b):  temp = temp.next  list1 = list2 # This doesn't work! We lose reference to return correct head

The Solution: Trust the algorithm - when a = 1:

• before_removal_node stays at the head (moved 0 times)
• after_removal_node moves to position b
• Setting before_removal_node.next = list2 correctly removes nodes 1 through b
• The original head is preserved and returned correctly

3. Not Handling Empty list2

The Pitfall: If list2 is empty (null), the traversal loop while before_removal_node.next: won't execute, and before_removal_node remains pointing to the node before position a.

Problematic Scenario:

# If list2 is None or empty: before_removal_node.next = list2 # Sets to None while before_removal_node.next: # Loop doesn't execute  before_removal_node = before_removal_node.next before_removal_node.next = after_removal_node.next # Still works correctly!

Why It Actually Works: The code handles this gracefully because when list2 is None, the connection still works properly - we're essentially just removing nodes without inserting anything.

Potential Enhancement (if needed):

if list2: # Only traverse if list2 exists  before_removal_node.next = list2  while before_removal_node.next:  before_removal_node = before_removal_node.next else:  # Direct connection when list2 is empty  pass # The next line handles it before_removal_node.next = after_removal_node.next

4. Confusion with 0-indexed vs 1-indexed Positions

The Pitfall: The problem statement uses 1-indexed positions (nodes are counted starting from 1), but the implementation treats a and b as if they follow this convention. Some developers might get confused and try to adjust for 0-indexing.

Wrong Adjustment:

# WRONG: Don't adjust if the problem already uses 1-indexing for _ in range(a - 2): # Incorrectly subtracting 2  before_removal_node = before_removal_node.next

The Solution: Carefully read whether the problem uses 0-indexing or 1-indexing. In this case:

• Position 1 = first node
• Position 2 = second node
• To get to the node before position a, move (a-1) times from the head