What is the time complexity of Merge In Between Linked Lists?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Merge In Between Linked Lists?

This problem uses Linked List. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Merge In Between Linked Lists?

Insight 1: We need to traverse `list1` to find the nodes at indices `a-1` and `b`. Insight 2: We need to find the tail of `list2` to connect it to the remaining part of `list1`.

Which companies ask LeetCode 1669. Merge In Between Linked Lists in interviews?

This problem is frequently asked at Arista Networks, PayPal.

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Merge In Between Linked Lists - LeetCode 1669 Solution

Merge In Between Linked Lists - Complete Solution Guide

Merge In Between Linked Lists is LeetCode problem 1669, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given two linked lists: list1 and list2 of sizes n and m respectively. Remove list1 's nodes from the a th node to the b th node, and put list2 in their place. The blue edges and nodes in the following figure indicate the result: Build the result list and return its head. Example 1: Input: list1 = [10,1,13,6,9,5], a = 3, b = 4, list2 = [1000000,1000001,1000002] Output: [10,1,13,1000000,1000001,1000002,5] Explanation: We remove the nodes 3 and 4 and put the entire list2 in their place. Th

Detailed Explanation

The problem asks us to modify `list1` by removing a sublist of nodes from index `a` to `b` (inclusive) and replacing it with `list2`. We need to return the modified `list1`. Specifically, the node at index `a-1` in `list1` should point to the head of `list2`, and the tail of `list2` should point to the node that originally followed index `b` in `list1`.

Solution Approach

The solution involves traversing `list1` to locate the node before the start of the sublist to be removed (`a-1`) and the node after the end of the sublist to be removed (`b+1`). Then, it connects the node at `a-1` to the head of `list2` and finds the tail of `list2`. Finally, it connects the tail of `list2` to the node originally at `b+1` in `list1`. This effectively splices `list2` into `list1` in place of the removed sublist.

Step-by-Step Algorithm

Step 1: Traverse `list1` from the head to the node at index `a-1`. This is done using a loop that iterates `a-1` times.
Step 2: Traverse `list1` from the node at `a-1` to the node at index `b`. Note that the code advances the pointer `b-a+1` times from the `a-1` node, effectively reaching the node indexed at `b`.
Step 3: Set the `next` pointer of the node at index `a-1` to point to the head of `list2`.
Step 4: Traverse `list2` to find its tail (the last node). This is done using a loop that continues until the `next` pointer is `None`.
Step 5: Set the `next` pointer of the tail of `list2` to point to the node that originally followed index `b` in `list1` (the node initially after `node_b`).
Step 6: Return the head of the modified `list1`.

Key Insights

Insight 1: We need to traverse `list1` to find the nodes at indices `a-1` and `b`.
Insight 2: We need to find the tail of `list2` to connect it to the remaining part of `list1`.
Insight 3: We need to handle the connections correctly to avoid breaking the list or creating cycles.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Linked List.

Companies

Asked at: Arista Networks, PayPal.

Merge In Between Linked Lists | LeetCode 1669

Medium

Linked List

Time: O(n)Space: O(1)

Solution Code

Available in:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeInBetween(ListNode list1, int a, int b, ListNode list2) {
        // Find the node at index a-1
        ListNode prev_a = list1;
        for (int i = 0; i < a - 1; i++) {
            prev_a = prev_a.next;
        }

        // Find the node at index b
        ListNode node_b = prev_a;
        for (int i = 0; i < b - a + 1; i++) {
            node_b = node_b.next;
        }

        // Connect the node at a-1 to list2's head
        prev_a.next = list2;

        // Find the tail of list2
        ListNode tail_list2 = list2;
        while (tail_list2.next != null) {
            tail_list2 = tail_list2.next;
        }

        // Connect list2's tail to the node at b+1
        tail_list2.next = node_b.next;

        return list1;
    }
}

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Problem Info

Description

Problem 1669: Merge In Between Linked Lists - Detailed explanation and solution approaches.

Topics

Linked List

Companies

Arista NetworksPayPal

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1669

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Traverse `list1` from the head to the node at index `a-1`. This is done using a loop that iterates `a-1` times.
2
Step 2: Traverse `list1` from the node at `a-1` to the node at index `b`. Note that the code advances the pointer `b-a+1` times from the `a-1` node, effectively reaching the node indexed at `b`.
3
Step 3: Set the `next` pointer of the node at index `a-1` to point to the head of `list2`.
4
Step 4: Traverse `list2` to find its tail (the last node). This is done using a loop that continues until the `next` pointer is `None`.
5
Step 5: Set the `next` pointer of the tail of `list2` to point to the node that originally followed index `b` in `list1` (the node initially after `node_b`).
6
Step 6: Return the head of the modified `list1`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the number of nodes we need to traverse. We traverse parts of list1 and the entirety of list2. The space complexity is constant because we are modifying the linked lists in-place without creating any new data structures that scale with the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using a Dummy Node

An alternative approach could use a dummy node at the beginning of `list1` to handle the edge case where `a` is 1 more cleanly. This eliminates a potential special case in the code.

Time: O(n + m)Space: O(1)

Trade-offs: The use of a dummy node can slightly improve readability in some cases, but it doesn't offer a significant performance benefit. It is often a matter of coding style preference.

✨

Key Insights

1Insight 1: We need to traverse `list1` to find the nodes at indices `a-1` and `b`.
2Insight 2: We need to find the tail of `list2` to connect it to the remaining part of `list1`.
3Insight 3: We need to handle the connections correctly to avoid breaking the list or creating cycles.

📚

Key Terms Explained

For Students

Linked List

Elements connected via pointers. Like a chain where each link knows the next.

Edge Cases

Edge case 1: `list2` is empty. The algorithm still works correctly because setting `prev_a.next = list2` will set it to `None`, effectively removing the sublist from `a` to `b`.
Edge case 2: `a` is 1 (we want to remove from the beginning of the list). The code will still work correctly, as it finds the node before the range to be removed. The first loop will not execute in this case.
Edge case 3: `b` is the last node in the list. The code works correctly by attaching the end of list2 to NULL.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to find the tail of `list2` before connecting it to the rest of `list1`. Connecting the head of `list2` directly to the rest of `list1` will leave the rest of list2 dangling.

✗Common mistake 2: Not handling the index calculations for `a` and `b` correctly, especially when `a` is 1. Incorrect index calculations can lead to connecting to the wrong nodes in `list1`.

✗Common mistake 3: Not considering the edge case where the list2 is empty, which can lead to null pointer exceptions if not handled correctly.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Text editors often use linked lists to represent text. This algorithm could be used to replace a range of text with a different string (represented as a linked list).

🏦Real-world scenario 2: In data processing pipelines, this could be used to replace a section of a linked list representing a sequence of operations with a new set of operations.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

The dummy node technique prevents edge cases. For cycle detection, Floyd's algorithm is the standard. Always handle empty list and single node cases.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Losing reference to nodes when modifying pointers. Use temporary variables to save references.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 20, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

How does the fast/slow pointer technique work?

Fast pointer moves 2 steps while slow moves 1 step. When fast reaches the end, slow is at middle. If there's a cycle, they will meet. This technique is used for: finding middle node, detecting cycles, finding cycle start point, and checking palindromes.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

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

For this medium-level Linked List problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?