What is the time complexity of Insert Delete GetRandom O(1)?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Insert Delete GetRandom O(1)?

This problem uses Array, Hash Table, Math, Design, Randomized. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Insert Delete GetRandom O(1)?

Insight 1: Using a hash map (dictionary) allows for O(1) average time complexity for `insert` and `remove` operations by providing quick lookups to check for the presence of an element and its index. Insight 2: Maintaining a list (array) allows for O(1) average time complexity for the `getRandom` operation since we can randomly access elements by index. However, removing an element from the middle of a list is typically O(n).

Which companies ask LeetCode 380. Insert Delete GetRandom O(1) in interviews?

This problem is frequently asked at Affirm, Agoda, AppFolio, Axon, ByteDance and 35 more companies.

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Insert Delete GetRandom O(1) - LeetCode 380 Solution

Insert Delete GetRandom O(1) - Complete Solution Guide

Insert Delete GetRandom O(1) is LeetCode problem 380, 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

Implement the RandomizedSet class: RandomizedSet() Initializes the RandomizedSet object. bool insert(int val) Inserts an item val into the set if not present. Returns true if the item was not present, false otherwise. bool remove(int val) Removes an item val from the set if present. Returns true if the item was present, false otherwise. int getRandom() Returns a random element from the current set of elements (it's guaranteed that at least one element exists when this method is called). Each ele

Detailed Explanation

The problem requires implementing a `RandomizedSet` data structure that supports three operations in O(1) average time complexity: inserting a value, removing a value, and retrieving a random value from the set. The `insert` operation should only add the value if it's not already present, returning `true` if inserted and `false` otherwise. The `remove` operation should only remove the value if it exists, returning `true` if removed and `false` otherwise. The `getRandom` operation should return a random element from the set, ensuring each element has an equal probability of being selected. The data structure must efficiently handle insertions, deletions, and random selection while maintaining O(1) average time complexity for all operations.

Solution Approach

The solution uses a combination of a hash map (dictionary) and a list (array). The hash map stores each value in the set as a key, with its corresponding index in the list as the value. The list stores all the values present in the set. `insert` checks if the value exists in the map; if not, it adds the value to the list and updates the map with the value and its index. `remove` checks if the value exists in the map; if so, it swaps the value to be removed with the last value in the list, updates the map, removes the last value from the list and removes the value from the map. `getRandom` generates a random index within the bounds of the list and returns the value at that index.

Step-by-Step Algorithm

Step 1: Initialize a hash map (dictionary) and a list (array) in the constructor of the `RandomizedSet` class.
Step 2: In the `insert` method, check if the value is already present in the hash map. If it is, return `false`.
Step 3: If the value is not present, add it to the list. Then, store the value and its index in the list into the hash map. Return `true`.
Step 4: In the `remove` method, check if the value is present in the hash map. If it isn't, return `false`.
Step 5: If the value is present, get its index from the hash map.
Step 6: Get the last element from the list.
Step 7: Replace the element to be removed with the last element.
Step 8: Update the hash map with the new index of the last element.
Step 9: Remove the last element from the list.
Step 10: Remove the value from the hash map. Return `true`.
Step 11: In the `getRandom` method, generate a random index within the bounds of the list's length.
Step 12: Return the element at the random index from the list.

Key Insights

Insight 1: Using a hash map (dictionary) allows for O(1) average time complexity for `insert` and `remove` operations by providing quick lookups to check for the presence of an element and its index.
Insight 2: Maintaining a list (array) allows for O(1) average time complexity for the `getRandom` operation since we can randomly access elements by index. However, removing an element from the middle of a list is typically O(n).
Insight 3: The key to achieving O(1) average time complexity for `remove` is to swap the element to be removed with the last element in the list and then pop the last element. This maintains the ability to randomly access elements in O(1) while avoiding shifting elements in the array.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Math, Design, Randomized.

Companies

Asked at: Affirm, Agoda, AppFolio, Axon, ByteDance, Cisco, Citadel, DE Shaw, Docusign, DoorDash, Goldman Sachs, Grammarly, Groupon, IXL, Intuit, LinkedIn, MakeMyTrip, Miro, Netflix, Nvidia, Okta, Palo Alto Networks, Peloton, Pocket Gems, Pure Storage, Quora, Rippling, Rubrik, Salesforce, Samsung, Snap, Snowflake, SoFi, Sprinklr, ThousandEyes, Unity, X, Yandex, Yelp, Zeta.

Insert Delete GetRandom O(1) | LeetCode 380

Medium

Array Hash Table Math Design Randomized

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

Solution Code

Available in:

import java.util.*;

class RandomizedSet {

    private Map<Integer, Integer> valMap;
    private List<Integer> valList;
    private Random random;

    public RandomizedSet() {
        valMap = new HashMap<>();
        valList = new ArrayList<>();
        random = new Random();
    }
    
    public boolean insert(int val) {
        if (valMap.containsKey(val)) {
            return false;
        }
        
        valMap.put(val, valList.size());
        valList.add(val);
        return true;
    }
    
    public boolean remove(int val) {
        if (!valMap.containsKey(val)) {
            return false;
        }
        
        int idxToRemove = valMap.get(val);
        int lastElement = valList.get(valList.size() - 1);
        
        valList.set(idxToRemove, lastElement);
        valMap.put(lastElement, idxToRemove);
        
        valList.remove(valList.size() - 1);
        valMap.remove(val);
        
        return true;
    }
    
    public int getRandom() {
        return valList.get(random.nextInt(valList.size()));
    }
}

/**
 * Your RandomizedSet object will be instantiated and called as such:
 * RandomizedSet obj = new RandomizedSet();
 * boolean param_1 = obj.insert(val);
 * boolean param_2 = obj.remove(val);
 * int param_3 = obj.getRandom();
 */

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Problem Info

Description

Problem 380: Insert Delete GetRandom O(1) - Detailed explanation and solution approaches.

Topics

Array Hash Table Math Design Randomized

Companies

AffirmAgodaAppFolioAxonByteDanceCiscoCitadelDE Shaw

Stats

Difficulty:Medium

Languages:4

Companies:40

Problem #:380

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

12 steps

1
Step 1: Initialize a hash map (dictionary) and a list (array) in the constructor of the `RandomizedSet` class.
2
Step 2: In the `insert` method, check if the value is already present in the hash map. If it is, return `false`.
3
Step 3: If the value is not present, add it to the list. Then, store the value and its index in the list into the hash map. Return `true`.
4
Step 4: In the `remove` method, check if the value is present in the hash map. If it isn't, return `false`.
5
Step 5: If the value is present, get its index from the hash map.
6
Step 6: Get the last element from the list.
7
Step 7: Replace the element to be removed with the last element.
8
Step 8: Update the hash map with the new index of the last element.
9
Step 9: Remove the last element from the list.
10
Step 10: Remove the value from the hash map. Return `true`.
11
Step 11: In the `getRandom` method, generate a random index within the bounds of the list's length.
12
Step 12: Return the element at the random index from the 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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(1) on average because hash map operations (insertion, deletion, lookup) take O(1) time on average. The list's random access is also O(1). The space complexity is O(n) since we store each element in both the list and hashmap.

📖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 balanced search tree (e.g., AVL tree, Red-Black tree)

A balanced search tree could be used to store the elements. This would maintain the elements in sorted order and allow for logarithmic time complexity for insert and remove operations.

Time: O(log n) for insert, remove, and getRandom (if we also maintain a list of elements for the getRandom operation).Space: O(n)

Trade-offs: Using a balanced search tree would provide a sorted set, which might be useful in some scenarios. However, the logarithmic time complexity for insert and remove operations is slower than the O(1) average time complexity achieved with the hash map and list approach. Therefore, this approach is only beneficial if sorted order is a requirement.

✨

Key Insights

1Insight 1: Using a hash map (dictionary) allows for O(1) average time complexity for `insert` and `remove` operations by providing quick lookups to check for the presence of an element and its index.
2Insight 2: Maintaining a list (array) allows for O(1) average time complexity for the `getRandom` operation since we can randomly access elements by index. However, removing an element from the middle of a list is typically O(n).
3Insight 3: The key to achieving O(1) average time complexity for `remove` is to swap the element to be removed with the last element in the list and then pop the last element. This maintains the ability to randomly access elements in O(1) while avoiding shifting elements in the array.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

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

Edge Cases

Edge case 1: Inserting a duplicate value. The code explicitly checks for the presence of the value in the hash map before inserting, preventing duplicates and returning `false` if the value already exists.
Edge case 2: Removing a non-existent value. The code checks for the presence of the value in the hash map before removing, returning `false` if the value does not exist.
Edge case 3: Calling getRandom on an empty set. While the problem statement guarantees that getRandom will not be called on an empty set, a robust implementation should include a check for this condition to avoid errors. (Note: the C code includes this check, while others do not)

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to update the hash map after swapping elements during removal. If the index of the last element in the list is not updated in the hash map after the swap, the hash map will contain incorrect index information, leading to incorrect removals.

✗Common mistake 2: Not handling the case where the element to be removed is the last element in the list. If the element to remove *is* the last, the code must not try to update the element moved into its place.

✗Common mistake 3: Incorrect random number generation. Using a biased random number generator can lead to uneven probability distribution in getRandom.

✅

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: Implementing a load balancer that distributes requests randomly across a set of servers. The `RandomizedSet` can be used to maintain the set of available servers and select one at random for each incoming request.

🏦Real-world scenario 2: Implementing a shuffling algorithm for a deck of cards or a playlist of songs. The `RandomizedSet` can be used to maintain the set of cards or songs and randomly select one for each iteration of the shuffling process.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 21, 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

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

@DevExpert92Jul 14

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

@AlgoMasterJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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 "Insert Delete GetRandom O(1)"?

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

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