What is the time complexity of Contains Duplicate II?

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

What data structures are used to solve Contains Duplicate II?

This problem uses Array, Hash Table, Sliding Window. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Contains Duplicate II?

Insight 1: Using a hash map (dictionary in Python) to store the numbers and their most recent indices allows for efficient lookup of duplicates within the `k` distance constraint. Insight 2: A sliding window approach is implicitly used. While not explicitly defined by two pointers, the `i - seen[num] <= k` condition implicitly checks the elements within a window of size `k` around each element.

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Contains Duplicate II - LeetCode 219 Solution

Q: How to solve Contains Duplicate II?

The provided code utilizes a hash map (or HashMap in Java) to efficiently track the most recent index of each number encountered in the array. It iterates through the array. For each number, it checks if the number is already in the hash map and if the difference between the current index and the previously seen index is less than or equal to `k`. If both conditions are true, it means a duplicate within the `k` distance has been found, and the function returns `true`. Otherwise, the current number's index is updated in the hash map, and the iteration continues. If the loop completes without finding such a duplicate, the function returns `false`.

Q: Which companies ask LeetCode 219. Contains Duplicate II in interviews?

This problem is frequently asked at Airbnb, Arista Networks, Goldman Sachs, Google, Netflix and 1 more companies.

Contains Duplicate II - Complete Solution Guide

Contains Duplicate II is LeetCode problem 219, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java.

Problem Statement

Given an integer array nums and an integer k , return true if there are two distinct indices i and j in the array such that nums[i] == nums[j] and abs(i - j) <= k . Example 1: Input: nums = [1,2,3,1], k = 3 Output: true Example 2: Input: nums = [1,0,1,1], k = 1 Output: true Example 3: Input: nums = [1,2,3,1,2,3], k = 2 Output: false Constraints: 1 <= nums.length <= 10 5 -10 9 <= nums[i] <= 10 9 0 <= k <= 10 5

Detailed Explanation

The problem asks whether an integer array contains any duplicate numbers within a specified distance `k` of each other. More formally, given an array `nums` and an integer `k`, we need to determine if there exist indices `i` and `j` such that `nums[i] == nums[j]` and `abs(i - j) <= k`. The function should return `true` if such a pair exists, and `false` otherwise. The input consists of the integer array `nums` and the integer `k`. The output is a boolean value indicating the presence or absence of the specified duplicate pair.

Solution Approach

The provided code utilizes a hash map (or HashMap in Java) to efficiently track the most recent index of each number encountered in the array. It iterates through the array. For each number, it checks if the number is already in the hash map and if the difference between the current index and the previously seen index is less than or equal to `k`. If both conditions are true, it means a duplicate within the `k` distance has been found, and the function returns `true`. Otherwise, the current number's index is updated in the hash map, and the iteration continues. If the loop completes without finding such a duplicate, the function returns `false`.

Step-by-Step Algorithm

Step 1: Initialize an empty hash map (dictionary in Python) called `seen` to store numbers and their indices.
Step 2: Iterate through the input array `nums` using a loop, accessing both the element and its index.
Step 3: For each number, check if the number is a key in `seen`. If it is, check if the absolute difference between the current index and the index stored in `seen` for that number is less than or equal to `k`.
Step 4: If the condition in Step 3 is true, return `true` because a duplicate within the specified distance is found.
Step 5: If the condition in Step 3 is false, update the `seen` map by storing the current index associated with that number.
Step 6: After iterating through all numbers, return `false` if no such duplicate pair was found.

Key Insights

Insight 1: Using a hash map (dictionary in Python) to store the numbers and their most recent indices allows for efficient lookup of duplicates within the `k` distance constraint.
Insight 2: A sliding window approach is implicitly used. While not explicitly defined by two pointers, the `i - seen[num] <= k` condition implicitly checks the elements within a window of size `k` around each element.
Insight 3: The solution avoids unnecessary comparisons by only checking for duplicates if a number is already present in the hash map. This significantly improves efficiency compared to a brute-force approach.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Sliding Window.

Companies

Asked at: Airbnb, Arista Networks, Goldman Sachs, Google, Netflix, Palantir Technologies.

Contains Duplicate II | LeetCode 219

Easy

Array Hash Table Sliding Window

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

Solution Code

Available in:

class Solution {
    public boolean containsNearbyDuplicate(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            if (map.containsKey(nums[i]) && i - map.get(nums[i]) <= k) {
                return true;
            }
            map.put(nums[i], i);
        }
        return false;
    }
}

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Problem Info

Description

Problem 219: Contains Duplicate II - Detailed explanation and solution approaches.

Topics

Array Hash Table Sliding Window

Companies

AirbnbArista NetworksGoldman SachsGoogleNetflixPalantir Technologies

Stats

Difficulty:Easy

Languages:2

Companies:6

Problem #:219

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: Initialize an empty hash map (dictionary in Python) called `seen` to store numbers and their indices.
2
Step 2: Iterate through the input array `nums` using a loop, accessing both the element and its index.
3
Step 3: For each number, check if the number is a key in `seen`. If it is, check if the absolute difference between the current index and the index stored in `seen` for that number is less than or equal to `k`.
4
Step 4: If the condition in Step 3 is true, return `true` because a duplicate within the specified distance is found.
5
Step 5: If the condition in Step 3 is false, update the `seen` map by storing the current index associated with that number.
6
Step 6: After iterating through all numbers, return `false` if no such duplicate pair was found.

💡

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 algorithm iterates through the input array once, resulting in a time complexity of O(n). The space complexity is determined by the hash map, which in the worst case stores all unique numbers up to `n` (if all numbers are unique). However, if k is smaller than n, only indices in a sliding window of size k will be stored, therefore making the space complexity O(min(n,k)).

📖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

Brute Force

Compare each element with all other elements within a distance of k using nested loops.

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

Trade-offs: This approach is less efficient than the hash map solution, especially for large arrays and larger values of `k`. It's simpler to understand but has a significantly higher time complexity.

✨

Key Insights

1Insight 1: Using a hash map (dictionary in Python) to store the numbers and their most recent indices allows for efficient lookup of duplicates within the `k` distance constraint.
2Insight 2: A sliding window approach is implicitly used. While not explicitly defined by two pointers, the `i - seen[num] <= k` condition implicitly checks the elements within a window of size `k` around each element.
3Insight 3: The solution avoids unnecessary comparisons by only checking for duplicates if a number is already present in the hash map. This significantly improves efficiency compared to a brute-force approach.

📚

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!

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: Empty array. The code handles this correctly; the loop won't execute, and `false` will be returned.
Edge case 2: `k` is 0. The code will correctly return `false` unless there are consecutive duplicates.
Edge case 3: All numbers are the same. The code will correctly identify this as a True case if k is at least 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the index difference calculation. Using `i - seen[num]` instead of `abs(i - seen[num])` might lead to incorrect results when the index of the previous occurrence is larger than the current index.

✗Common mistake 2: Forgetting to update the hash map with the current index after checking for duplicates. This will lead to incorrect results in scenarios with repeated numbers.

✅

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: Network security. Detecting repeated patterns (e.g., malicious network traffic) within a certain time window.

🏦Real-world scenario 2: Data stream analysis. Identifying duplicate entries in a continuously flowing data stream within a defined time frame.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

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

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 10, 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 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Contains Duplicate II"?

For this easy-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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