What is the time complexity of Remove Zero Sum Consecutive Nodes from Linked List?

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

What data structures are used to solve Remove Zero Sum Consecutive Nodes from Linked List?

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

What are the key insights for solving Remove Zero Sum Consecutive Nodes from Linked List?

Insight 1: Using prefix sums allows us to quickly identify sublists that sum to zero. If the prefix sum at two different nodes is the same, the sublist between those nodes sums to zero. Insight 2: A hash table can be used to store the prefix sums and their corresponding nodes, enabling efficient lookups to identify the end node of a zero-sum sublist.

Which companies ask LeetCode 1171. Remove Zero Sum Consecutive Nodes from Linked List in interviews?

This problem is frequently asked at ByteDance, josh technology.

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Remove Zero Sum Consecutive Nodes from Linked List - LeetCode 1171 Solution

Remove Zero Sum Consecutive Nodes from Linked List - Complete Solution Guide

Remove Zero Sum Consecutive Nodes from Linked List is LeetCode problem 1171, 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

Given the head of a linked list, we repeatedly delete consecutive sequences of nodes that sum to 0 until there are no such sequences. After doing so, return the head of the final linked list. You may return any such answer. (Note that in the examples below, all sequences are serializations of ListNode objects.) Example 1: Input: head = [1,2,-3,3,1] Output: [3,1] Note: The answer [1,2,1] would also be accepted. Example 2: Input: head = [1,2,3,-3,4] Output: [1,2,4] Example 3: Input: head = [1,2,3,

Detailed Explanation

The problem asks us to remove all consecutive sequences of nodes in a linked list that sum to zero. We should repeatedly perform this removal until no such sequences remain. The goal is to return the head of the modified linked list after all zero-sum subsequences have been removed. For example, given the linked list [1, 2, -3, 3, 1], the subsequence [1, 2, -3] sums to zero, leaving [3, 1]. However, the problem statement notes that [1, 2, 1] would also be accepted. Given [1, 2, 3, -3, 4], the subsequence [3, -3] sums to zero, leaving [1, 2, 4]. Finally, given [1, 2, 3, -3, -2], the subsequence [2, 3, -3, -2] sums to zero, leaving [1].

Solution Approach

The solution uses a hash table to store the prefix sums encountered while traversing the linked list. For each prefix sum, the corresponding node is stored in the hash table. If a prefix sum is encountered again, it means that the nodes between the previous occurrence of the prefix sum and the current node sum to zero. In this case, we update the `next` pointer of the node corresponding to the previous occurrence of the prefix sum to point to the node after the current node, effectively removing the zero-sum sublist. This process is repeated until the end of the list is reached. A dummy node is used at the beginning to simplify the handling of the list's head.

Step-by-Step Algorithm

Step 1: Create a dummy node and set its `next` pointer to the head of the linked list. This simplifies handling cases where the beginning of the list forms a zero-sum sublist.
Step 2: Initialize a hash table to store prefix sums and their corresponding nodes. Initialize `prefix_sum` to 0.
Step 3: Traverse the linked list, starting from the dummy node. In each iteration, update `prefix_sum` by adding the current node's value.
Step 4: Store the `prefix_sum` and the current node in the hash table. If the `prefix_sum` already exists in the hash table, it indicates that a zero-sum sublist exists between the node previously stored and the current node.
Step 5: After the first traversal to populate the hashmap, reset `prefix_sum` to 0 and traverse the list again. This time, for each node, update the `next` pointer of the current node to point to the node after the zero-sum sublist (which is found using the stored value in the HashMap associated with the prefix sum).
Step 6: Return the `next` pointer of the dummy node, which will be the head of the modified linked list.

Key Insights

Insight 1: Using prefix sums allows us to quickly identify sublists that sum to zero. If the prefix sum at two different nodes is the same, the sublist between those nodes sums to zero.
Insight 2: A hash table can be used to store the prefix sums and their corresponding nodes, enabling efficient lookups to identify the end node of a zero-sum sublist.
Insight 3: A dummy node simplifies the handling of edge cases, such as when the head of the list is part of a zero-sum sequence.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, Linked List.

Companies

Asked at: ByteDance, josh technology.

Remove Zero Sum Consecutive Nodes from Linked List | LeetCode 1171

Medium

Hash Table Linked List

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

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; }
 * }
 */
import java.util.HashMap;

class Solution {
    public ListNode removeZeroSumSublists(ListNode head) {
        ListNode dummy = new ListNode(0, head);
        int prefix_sum = 0;
        HashMap<Integer, ListNode> seen = new HashMap<>();
        
        ListNode current = dummy;
        while (current != null) {
            prefix_sum += current.val;
            seen.put(prefix_sum, current);
            current = current.next;
        }
        
        prefix_sum = 0;
        current = dummy;
        while (current != null) {
            prefix_sum += current.val;
            current.next = seen.get(prefix_sum).next;
            current = current.next;
        }
        
        return dummy.next;
    }
}

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Problem Info

Description

Problem 1171: Remove Zero Sum Consecutive Nodes from Linked List - Detailed explanation and solution approaches.

Topics

Hash Table Linked List

Companies

ByteDancejosh technology

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1171

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: Create a dummy node and set its `next` pointer to the head of the linked list. This simplifies handling cases where the beginning of the list forms a zero-sum sublist.
2
Step 2: Initialize a hash table to store prefix sums and their corresponding nodes. Initialize `prefix_sum` to 0.
3
Step 3: Traverse the linked list, starting from the dummy node. In each iteration, update `prefix_sum` by adding the current node's value.
4
Step 4: Store the `prefix_sum` and the current node in the hash table. If the `prefix_sum` already exists in the hash table, it indicates that a zero-sum sublist exists between the node previously stored and the current node.
5
Step 5: After the first traversal to populate the hashmap, reset `prefix_sum` to 0 and traverse the list again. This time, for each node, update the `next` pointer of the current node to point to the node after the zero-sum sublist (which is found using the stored value in the HashMap associated with the prefix sum).
6
Step 6: Return the `next` pointer of the dummy node, which will be 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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because we iterate through the linked list twice. The space complexity is O(n) because, in the worst case, we store each prefix sum in the hashmap. For instance, the linked list [1, 2, 3, 4, 5] will store all prefix sums up to 15.

📖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

Two Pointers

Iterate through all possible sublists and check their sum. Remove if the sum is zero. This approach does not require additional space, but it is less efficient.

Time: O(n^3) in the worst case (n^2 to iterate through all sublists, and O(n) to compute the sum of each sublist).Space: O(1)

Trade-offs: The two-pointer approach is simpler to implement without additional data structures, but it has a much higher time complexity, making it unsuitable for large lists. The hashmap approach offers optimal time complexity at the cost of linear space.

✨

Key Insights

1Insight 1: Using prefix sums allows us to quickly identify sublists that sum to zero. If the prefix sum at two different nodes is the same, the sublist between those nodes sums to zero.
2Insight 2: A hash table can be used to store the prefix sums and their corresponding nodes, enabling efficient lookups to identify the end node of a zero-sum sublist.
3Insight 3: A dummy node simplifies the handling of edge cases, such as when the head of the list is part of a zero-sum sequence.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Linked List

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

Edge Cases

Edge case 1: The entire list sums to zero. The solution correctly handles this by removing the entire list.
Edge case 2: The head of the list is part of a zero-sum sequence. The dummy node ensures that the head can be correctly removed if necessary.
Edge case 3: Multiple overlapping zero-sum sequences. The hash table ensures correct removal even with overlaps.
Edge case 4: Empty List. The dummy node handles it gracefully, returning null.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using a dummy node. This makes handling cases where the head of the list needs to be removed more complex.

✗Common mistake 2: Not properly updating the `next` pointers after identifying a zero-sum sublist. This can lead to incorrect removal or cycles in the list.

✗Common mistake 3: Assuming that a single pass is sufficient to remove all zero-sum sublists. The removal process may create new zero-sum sublists, so repeated passes might be necessary. The provided solution makes correct updates to next pointers which resolves the need for repeated calls.

✗Common mistake 4: Incorrectly computing or handling prefix sums, potentially leading to inaccurate removal of sublists.

✅

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 cleaning - Removing sequences of transactions that cancel each other out in a financial ledger.

🏦Real-world scenario 2: Signal processing - Filtering out noise that has a zero-sum characteristic from a signal stream.

🎮Real-world scenario 3: Compiler Design - Simplifying or optimizing code by removing sections that effectively have no net effect (sum to zero) on the program's state.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

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

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

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: August 31, 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 do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

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 "Remove Zero Sum Consecutive Nodes from Linked List"?

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

🤖 Need help with this problem?