What is the time complexity of Merge Nodes in Between Zeros?

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

What data structures are used to solve Merge Nodes in Between Zeros?

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

What are the key insights for solving Merge Nodes in Between Zeros?

Insight 1: The linked list structure is already provided, we just need to traverse and modify it in place. Insight 2: The problem can be solved with a single pass through the linked list.

Which companies ask LeetCode 2181. Merge Nodes in Between Zeros in interviews?

This problem is frequently asked at josh technology.

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Merge Nodes in Between Zeros - LeetCode 2181 Solution

Merge Nodes in Between Zeros - Complete Solution Guide

Merge Nodes in Between Zeros is LeetCode problem 2181, 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 the head of a linked list, which contains a series of integers separated by 0 's. The beginning and end of the linked list will have Node.val == 0 . For every two consecutive 0 's, merge all the nodes lying in between them into a single node whose value is the sum of all the merged nodes. The modified list should not contain any 0 's. Return the head of the modified linked list . Example 1: Input: head = [0,3,1,0,4,5,2,0] Output: [4,11] Explanation: The above figure represents the

Detailed Explanation

The problem requires merging nodes between consecutive zeros in a linked list. The input is a linked list guaranteed to start and end with a node with value 0, and there are no consecutive nodes with value 0 in between. The task is to replace the nodes between each pair of consecutive zeros with a single node whose value is the sum of all nodes between the zeros. Finally, the list shouldn't contain any zeros other than the implied start and finish zero.

Solution Approach

The provided solution uses a single-pass approach to traverse the linked list. It maintains a 'current_sum' to accumulate the values between zeros. When a zero is encountered, the accumulated sum is assigned to the 'next' node of the 'tail' pointer. The 'tail' pointer is then advanced to this newly modified node. Finally, after processing all nodes, the 'next' pointer of the last valid node ('tail') is set to 'None' to terminate the list correctly.

Step-by-Step Algorithm

Step 1: Initialize 'tail' to 'head', 'current' to 'head.next', and 'current_sum' to 0.
Step 2: Iterate through the linked list using the 'current' pointer until 'current' is None.
Step 3: If 'current.val' is not 0, add 'current.val' to 'current_sum'.
Step 4: If 'current.val' is 0, assign 'current_sum' to 'tail.next.val', move 'tail' to 'tail.next', and reset 'current_sum' to 0.
Step 5: Move 'current' to 'current.next'.
Step 6: After the loop finishes, set 'tail.next' to None to remove any trailing nodes.
Step 7: Return 'head.next' as the head of the modified linked list.

Key Insights

Insight 1: The linked list structure is already provided, we just need to traverse and modify it in place.
Insight 2: The problem can be solved with a single pass through the linked list.
Insight 3: Using a 'tail' pointer to keep track of the last modified node simplifies the in-place modification.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Linked List, Simulation.

Companies

Asked at: josh technology.

Merge Nodes in Between Zeros | LeetCode 2181

Medium

Linked List Simulation

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 mergeNodes(ListNode head) {
        ListNode tail = head;
        ListNode current = head.next;
        int current_sum = 0;
        
        while (current != null) {
            if (current.val != 0) {
                current_sum += current.val;
            } else {
                tail.next.val = current_sum;
                tail = tail.next;
                current_sum = 0;
            }
            
            current = current.next;
        }
        
        tail.next = null;
        
        return head.next;
    }
}

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Problem Info

Description

Problem 2181: Merge Nodes in Between Zeros - Detailed explanation and solution approaches.

Topics

Linked List Simulation

Companies

josh technology

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2181

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

7 steps

1
Step 1: Initialize 'tail' to 'head', 'current' to 'head.next', and 'current_sum' to 0.
2
Step 2: Iterate through the linked list using the 'current' pointer until 'current' is None.
3
Step 3: If 'current.val' is not 0, add 'current.val' to 'current_sum'.
4
Step 4: If 'current.val' is 0, assign 'current_sum' to 'tail.next.val', move 'tail' to 'tail.next', and reset 'current_sum' to 0.
5
Step 5: Move 'current' to 'current.next'.
6
Step 6: After the loop finishes, set 'tail.next' to None to remove any trailing nodes.
7
Step 7: Return 'head.next' as the head of the modified linked list.

💡

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 linear because each node in the linked list is visited exactly once. The space complexity is constant because the algorithm operates in place without using any additional 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

Creating a New Linked List

Instead of modifying the list in-place, we could create a new linked list to store the merged node values. This approach would iterate through the original list, calculate the sums between zeros, and create new nodes with these sums in the new linked list.

Time: O(n)Space: O(m), where m is the number of merged nodes (number of zeros - 1).

Trade-offs: This approach uses extra space to create a new list but might be preferred if the original list needs to be preserved. In-place modification is more memory-efficient if the original list can be altered.

✨

Key Insights

1Insight 1: The linked list structure is already provided, we just need to traverse and modify it in place.
2Insight 2: The problem can be solved with a single pass through the linked list.
3Insight 3: Using a 'tail' pointer to keep track of the last modified node simplifies the in-place modification.

📚

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: Input list is [0, x, 0], where x is some value. The algorithm correctly sums the value of 'x' and the final modified list becomes [x].
Edge case 2: The given problem states there is no consecutive zeros and it starts and ends with zero node. Thus we don't need to worry about handling such cases.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not setting 'tail.next' to None at the end. This can result in a linked list that doesn't terminate correctly, potentially leading to infinite loops or incorrect results in subsequent operations.

✗Common mistake 2: Incorrectly updating the 'tail' pointer. The 'tail' should only be moved forward when a zero is encountered, after the sum is assigned to the next node of the previous 'tail'. Moving 'tail' at other times will break the linked list structure.

✗Common mistake 3: Forgetting to return 'head.next' because head is 0 in the first place.

✅

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: Data processing where consecutive data points need to be aggregated (e.g., summing daily sales figures into monthly totals). The linked list could represent the daily sales, and the zeros could represent the end of a month.

🏦Real-world scenario 2: Packet processing in networking, where packets belonging to the same session (delimited by special markers like zeros) need to be aggregated for further analysis or forwarding.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: June 22, 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

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

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 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 Nodes in Between Zeros"?

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 simulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?