What is the time complexity of Linked List Random Node?

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

What data structures are used to solve Linked List Random Node?

This problem uses Linked List, Math, Reservoir Sampling, Randomized. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Linked List Random Node?

Insight 1: Reservoir Sampling is a technique useful for selecting a random element from a stream of data of unknown length. In this case, the linked list can be treated as a stream of data. Insight 2: To ensure each node has an equal probability of being selected, the probability of selecting the i-th node should be 1/i after processing i nodes.

Which companies ask LeetCode 382. Linked List Random Node in interviews?

This problem is frequently asked at Nvidia.

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Linked List Random Node - LeetCode 382 Solution

Linked List Random Node - Complete Solution Guide

Linked List Random Node is LeetCode problem 382, 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 a singly linked list, return a random node's value from the linked list. Each node must have the same probability of being chosen. Implement the Solution class: Solution(ListNode head) Initializes the object with the head of the singly-linked list head . int getRandom() Chooses a node randomly from the list and returns its value. All the nodes of the list should be equally likely to be chosen. Example 1: Input ["Solution", "getRandom", "getRandom", "getRandom", "getRandom", "getRandom"] [[

Detailed Explanation

The problem requires us to implement a class that can randomly select a node's value from a singly linked list, with each node having an equal probability of being chosen. The `Solution` class has a constructor that takes the head of the linked list and a `getRandom()` method that returns the value of a randomly selected node. The key constraint is that each node must have the same probability of being selected.

Solution Approach

The solution utilizes Reservoir Sampling to achieve equal probability selection without knowing the length of the linked list beforehand. We iterate through the linked list, and at each node, we have a probability of 1/count to replace the current result with the node's value, where 'count' is the number of nodes visited so far. This ensures that each node has an equal probability of being selected after traversing the entire list.

Step-by-Step Algorithm

Step 1: Initialize `count` to 0 and `result` to 0. These variables will track the number of nodes seen so far and the value of the randomly selected node, respectively.
Step 2: Traverse the linked list starting from the head.
Step 3: For each node encountered, increment the `count` by 1.
Step 4: Generate a random integer between 1 and `count` (inclusive).
Step 5: If the generated random integer is equal to 1 (probability of 1/count), update the `result` to the current node's value.
Step 6: Move to the next node in the linked list and repeat steps 3-5 until the end of the list is reached.
Step 7: Return the `result`.

Key Insights

Insight 1: Reservoir Sampling is a technique useful for selecting a random element from a stream of data of unknown length. In this case, the linked list can be treated as a stream of data.
Insight 2: To ensure each node has an equal probability of being selected, the probability of selecting the i-th node should be 1/i after processing i nodes.
Insight 3: The solution must operate in O(1) space and be able to handle very large linked lists where pre-computing the list length is undesirable or impossible.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Linked List, Math, Reservoir Sampling, Randomized.

Companies

Asked at: Nvidia.

Linked List Random Node | LeetCode 382

Medium

Linked List Math Reservoir Sampling Randomized

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

Solution Code

Available in:

import java.util.Random;

/**
 * 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 {

    private ListNode head;
    private Random random;

    public Solution(ListNode head) {
        this.head = head;
        this.random = new Random();
    }
    
    public int getRandom() {
        int count = 0;
        int result = 0;
        ListNode current = this.head;
        
        while (current != null) {
            count++;
            // With probability 1/count, select the current node's value
            if (random.nextInt(count) == 0) { //nextInt(count) generates a number between 0 (inclusive) and count (exclusive). So, == 0 gives 1/count probability
                result = current.val;
            }
            current = current.next;
        }
        
        return result;
    }
}

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(head);
 * int param_1 = obj.getRandom();
 */

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Problem Info

Description

Problem 382: Linked List Random Node - Detailed explanation and solution approaches.

Topics

Linked List Math Reservoir Sampling Randomized

Companies

Nvidia

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:382

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 `count` to 0 and `result` to 0. These variables will track the number of nodes seen so far and the value of the randomly selected node, respectively.
2
Step 2: Traverse the linked list starting from the head.
3
Step 3: For each node encountered, increment the `count` by 1.
4
Step 4: Generate a random integer between 1 and `count` (inclusive).
5
Step 5: If the generated random integer is equal to 1 (probability of 1/count), update the `result` to the current node's value.
6
Step 6: Move to the next node in the linked list and repeat steps 3-5 until the end of the list is reached.
7
Step 7: Return the `result`.

💡

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 O(n) because we need to traverse the entire linked list once in the getRandom() function. The space complexity is O(1) as we are only using a few integer variables and a pointer, which takes constant space regardless of 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

Storing the Linked List in an Array

Convert the linked list into an array or ArrayList. Then, generate a random index within the bounds of the array and return the element at that index.

Time: O(n) for conversion, O(1) for getRandom()Space: O(n)

Trade-offs: This approach has O(1) time complexity for getRandom() after the initial conversion but requires O(n) extra space to store the entire list. This is inefficient if memory is a constraint and getRandom() is called very frequently.

✨

Key Insights

1Insight 1: Reservoir Sampling is a technique useful for selecting a random element from a stream of data of unknown length. In this case, the linked list can be treated as a stream of data.
2Insight 2: To ensure each node has an equal probability of being selected, the probability of selecting the i-th node should be 1/i after processing i nodes.
3Insight 3: The solution must operate in O(1) space and be able to handle very large linked lists where pre-computing the list length is undesirable or impossible.

📚

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: Empty linked list. The code handles this correctly as the while loop will not execute and it will implicitly return a default value of result (0 in Python and Java). However, the problem statement specifies that the list will always have at least one node.
Edge case 2: Single node linked list. The count will be 1, the random number will always be 1, and the result will be the value of that single node with 100% probability, which is the expected behavior.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using `random.random()` instead of `random.randint(1, count)` or `random.nextInt(count)` which does not give each element an equal probability of being selected.

✗Common mistake 2: Attempting to pre-compute the length of the linked list and then selecting a random index. This requires an extra traversal of the linked list, increasing the time complexity. Also, in the follow-up case where the list is extremely long or streaming, this pre-computation is not feasible.

✅

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: Load balancing in a distributed system. Imagine a server needs to randomly select a worker node from a dynamic pool of available nodes, where nodes can join or leave the pool at any time. Reservoir sampling can be used to maintain a uniform random selection without needing to know the total number of worker nodes.

🏦Real-world scenario 2: Data stream processing. When analyzing a stream of data, such as network traffic or sensor readings, we might want to randomly sample data points to reduce the dataset size for analysis. Reservoir sampling enables us to achieve this with equal probability for each data point without storing the entire stream in memory.

💡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: June 23, 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

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

@InterviewProJul 13

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

@CodeNinjaJul 6

❓

Frequently Asked Questions

Why use a dummy node in linked list problems?

A dummy node simplifies edge cases where the head might change (insertion/deletion at head) or when building a new list. Initialize dummy.next = head, work with dummy.next throughout, and return dummy.next at the end to handle all cases uniformly.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Linked List Random Node"?

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

🤖 Need help with this problem?