1669. Merge In Between Linked Lists - Detailed Explanation

Problem Statement

You are given two linked lists: list1 and list2. Also, you are given two integers, a and b. Your task is to remove the nodes in list1 from index a to b (inclusive) and then insert list2 in their place. Finally, return the head of the modified list.

Example Inputs and Outputs

Example 1:

• Input:
  • list1: 0 → 1 → 2 → 3 → 4 → 5
  • list2: 1000000 → 1000001 → 1000002
  • a = 3, b = 4
• Output: 0 → 1 → 2 → 1000000 → 1000001 → 1000002 → 5
• Explanation:
  Nodes 3 and 4 in list1 (at indices 3 and 4) are removed. list2 is inserted between node 2 and node 5.

Example 2:

• Input:
  • list1: 0 → 1 → 2 → 3 → 4 → 5 → 6
  • list2: 1000000 → 1000001 → 1000002 → 1000003 → 1000004 → 1000005
  • a = 2, b = 5
• Output: 0 → 1 → 1000000 → 1000001 → 1000002 → 1000003 → 1000004 → 1000005 → 6
• Explanation:
  Nodes 2, 3, 4, and 5 (indices 2 through 5) are removed from list1. list2 is inserted between node 1 and node 6.

Example 3:

• Input:
  • list1: 5 → 6 → 7 → 8 → 9
  • list2: 1 → 2 → 3
  • a = 1, b = 2
• Output: 5 → 1 → 2 → 3 → 8 → 9
• Explanation:
  Nodes 6 and 7 (indices 1 and 2) in list1 are removed and list2 is inserted between node 5 and node 8.

Constraints:

• The indices a and b will always be valid (0-indexed) and a <= b.
• list1 will have at least b + 1 nodes.
• list2 is a non-empty linked list.
• The problem assumes you have basic operations on linked lists (traversal, pointer manipulation).

Hints Before the Solution

1. Find Key Pointers in list1:

   • Traverse list1 to find the node immediately before the ath node (let’s call it nodeA_prev).
   • Also, find the node at index b (let’s call it nodeB). You will need the node right after nodeB to connect to list2 later.

2. Find the Tail of list2:

   • Traverse list2 to find its last node (the tail). This tail will eventually be connected to the node that comes after nodeB in list1.

3. Connect the Lists:

   • Set nodeA_prev.next to the head of list2.
   • Then, set the tail of list2.next to the node after nodeB.

Approaches

Approach 1: Brute Force Rebuild

In the brute force approach, instead of modifying the original list in place, you build a new linked list from scratch by copying over the nodes you want to keep and then inserting the nodes from the second list. This approach is straightforward conceptually, especially if you’re not comfortable with pointer manipulations.

Steps:

1. Traverse and Copy First Part of list1:

   • Start from the head of list1.
   • Traverse until you reach index a - 1 (i.e., the node just before the removal section).
   • For each node up to index a - 1, create a copy (or just reassign if allowed) and append it to your new list.

2. Insert All Nodes from list2:

   • Traverse list2 from its head.
   • Append each node from list2 to the new list.
   • Note: This requires iterating through the entire list2 to get all its nodes.

3. Traverse and Copy the Remaining Part of list1:

   • Continue from index b + 1 in list1.
   • Copy all remaining nodes from list1 into the new list.

4. Return the Head of the New List:

   • The new list now represents the merged result with the nodes from index a to b removed and list2 inserted in their place.

Pros and Cons:

• Pros:

  • Simplicity: The logic of building a new list is straightforward.
  • Clarity: You avoid complex pointer manipulations that can lead to bugs.

• Cons:

  • Extra Space: You use additional space for the new list (even if you reuse node values).
  • Inefficiency: The need to traverse parts of list1 twice (once to copy the beginning and once for the ending) and the whole of list2 increases time overhead compared to a direct pointer reattachment.

When to Use This Approach:

• When you want to prioritize code clarity over efficiency.
• In environments where extra space isn’t a concern, or when working in languages/environments that make pointer manipulations error-prone.

Approach 2: Optimal In‑Place Merge

This approach focuses on adjusting the pointers in the original lists rather than building a new list. The idea is to splice list2 directly into list1 by removing a section of list1 and reconnecting the nodes without creating new nodes or using extra memory.

Detailed Steps:

1. Find the Node Before the Removal Section (nodeA_prev):

   • Traverse list1 from the head.
   • Stop when you reach the node at index a - 1 (this node will serve as the connecting point for list2).
   • Visualization:
     If list1 is 0 → 1 → 2 → 3 → 4 → 5 and a = 3, then nodeA_prev is the node with value 2.

2. Locate the End of the Removal Section (nodeB):

   • Continue from nodeA_prev (or start a fresh traversal) until you reach the node at index b.
   • This node marks the end of the section that will be removed.
   • Visualization:
     For the same list1 with b = 4, nodeB is the node with value 4.

3. Find the Tail of list2:

   • Traverse list2 to get to its last node.
   • This tail will later be used to connect back to list1.
   • Visualization:
     If list2 is 1000000 → 1000001 → 1000002, then the tail is the node with value 1000002.

4. Reconnect Pointers to Merge the Lists:

   • Linking the Start:
     Set the next pointer of nodeA_prev to point to the head of list2.
   • Linking the End:
     Set the next pointer of the tail of list2 to the node that comes immediately after nodeB in list1 (i.e., nodeB.next).
   • Result:
     This effectively removes the nodes from index a to b in list1 and inserts list2 in their place without using extra space.

5. Return the Modified list1:

   • The head of list1 remains unchanged (unless a is 0, which would require updating the head).

Pros and Cons:

• Pros:

  • Space Efficient: Uses only a few pointers without allocating extra memory.
  • Time Efficient: Each list is traversed only as much as needed.
  • Direct: Directly manipulates the original data structure, which is often required in interview settings.

• Cons:

  • Pointer Complexity: Requires careful pointer manipulation. A small mistake can lead to memory leaks or broken links between nodes.
  • Edge Cases: Special cases (e.g., when a is 0) must be handled explicitly, which may complicate the logic.

When to Use This Approach:

• In interviews and real-world scenarios where space and time efficiency are paramount.
• When you are comfortable with linked list pointer manipulations and want to optimize for performance.