What is the brute force approach and complexity for Merge In Between Linked Lists?

Put the nodes of both lists in vectors, build the new order, relink with time complexity O(N + M) and space complexity O(N + M).

What is the optimal approach and complexity for Merge In Between Linked Lists?

Walk to node a − 1 and node b, splice list2 in, link its tail to node b + 1 with optimal time complexity O(N + M) and space complexity O(1).

How do you identify the pattern for Merge In Between Linked Lists?

Look for key signals in the prompt: "replace positions a to b" indicates A block in the middle is removed and another list goes there → find node a − 1 and node b, then fix only the two ends.; "list2 goes in as a whole" indicates Splice the chain: node a − 1 points at list2's head, and list2's tail points at node b + 1. No values are compared.; "singly linked list" indicates Save the link to node b + 1 in a variable before changing any `next`..

Merge In Between Linked Lists

Medium·LeetCode #1669 Pro OnlySlight Variation

LeetCode

Problem Statement

You get two linked lists, list1 and list2. Positions in list1 start at 0.

Remove the nodes of list1 from position a to position b, both ends included. Then put all of list2 in their place. Afterwards, list2 sits between the node at position a − 1 and the node that was at position b + 1. Return the head of list1.

Input

list1 and list2, plus two positions a and b (with a ≤ b) that refer to list1.

Output

The head of list1 after the nodes from a to b are replaced by list2.

Example 1

Input: list1 = [0, 1, 2, 3, 4, 5], a = 3, b = 4, list2 = [1000000, 1000001, 1000002, 1000003, 1000004]

Output: [0, 1, 2, 1000000, 1000001, 1000002, 1000003, 1000004, 5]

Positions 3 and 4 (values 3 and 4) are removed. list2 goes in their place, and the 5 stays at the end.

Example 2

Input: list1 = [0, 1, 2, 3, 4, 5, 6], a = 2, b = 5, list2 = [1000000, 1000001, 1000002, 1000003, 1000004]

Output: [0, 1, 1000000, 1000001, 1000002, 1000003, 1000004, 6]

Positions 2 to 5 (values 2, 3, 4, 5) are removed. The 0 and the 1 stay before list2, and the 6 stays after it.

Constraints

• list1 has between 3 and 10^4 nodes.
• list2 has between 1 and 10^4 nodes.
• 1 ≤ a ≤ b < n − 1, where n is the number of nodes in list1. So there is always a node before the gap and a node after it.
• The node values are between 0 and 10^6.

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“replace positions a to b”

→ A block in the middle is removed and another list goes there → find node a − 1 and node b, then fix only the two ends.

“list2 goes in as a whole”

→ Splice the chain: node a − 1 points at list2's head, and list2's tail points at node b + 1. No values are compared.

“singly linked list”

→ Save the link to node b + 1 in a variable before changing any `next`.

Approach Complexity Ladder

Brute Force — collect the nodes in arrays, then relinkO(N + M)O(N + M)

Put the nodes of both lists in vectors, build the new order, relink

Optimal — walk to the gap, splice list2 in, reconnect the tailO(N + M)O(1)

Walk to node a − 1 and node b, splice list2 in, link its tail to node b + 1

Brute Force Intuition

Changing links in the middle of a list is easy to get wrong. A vector makes the new order obvious. Put the nodes of list1 and of list2 into two vectors. The new order is the first a nodes of list1, then all of list2, then the nodes after position b. Link each node to the next one in that order.

Recognition rule for this sub-pattern:

Find the node before the range, reverse exactly the range, then stitch both ends back.

Remember this · Linked List — Reversal

Trigger

Reverse whole / part / every k nodes, reorder, palindrome check, numbers stored as lists.

Think

Which piece do I reverse, and which nodes must I remember to reconnect afterwards?

Key question

Before reversing, who is the node just before the piece and who is just after?

Brute2Best Pro Content

Unlock Complete Visual Walkthrough & Optimal C++

The interactive animated dry-run player, optimized C++ implementation, state invariant derivations, and edge-case breakdown for this question are part of Brute2Best Pro. The anchor question of every sub-pattern is completely free.

All 440 visual dry runs & algorithms

Interactive step-by-step playback with pointer & array state transitions

52 comprehensive LLD chapters

Interactive UML diagrams, real-world stories (Food Delivery, Trading, RPG), and practice solutions

40 C++ basics labs & hardware memory models

Vtables, byte layouts, assembly insights, and multi-threading race conditions

Production-ready C++23 source code

Modern idioms, clean contracts, time/space complexity derivations, and interview takeaways

Create Free Account Log in

First anchor question per pattern, Part 1 LLD & Module 1 C++ basics are 100% Free forever.

Beyond The LeetCode Problem

How this concept connects to System Design & C++ Systems

Bridge DSA with production engineering

Top engineering interviews test whether you understand how algorithms like Linked List — Reversal translate into production architectures and low-level memory mechanics.

LLD Architecture