What is the time complexity of Contains Duplicate III?

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

What data structures are used to solve Contains Duplicate III?

This problem uses Array, Sliding Window, Sorting, Bucket Sort, Ordered Set. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Contains Duplicate III?

Insight 1: Using a sliding window approach and checking all pairs within the window will result in O(n*k) time complexity (where 'k' is indexDiff), which can be too slow for large input sizes. The constraints indicate that an O(n) solution is expected. Insight 2: The bucket sort/pigeonhole principle helps group nearby values into buckets, enabling efficient comparison of elements close in value. By using a bucket size of `valueDiff + 1`, any two elements within the same bucket or adjacent buckets must have a value difference less than or equal to `valueDiff`.

Which companies ask LeetCode 220. Contains Duplicate III in interviews?

This problem is frequently asked at Airbnb, Netflix, Palantir Technologies, Yandex.

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Contains Duplicate III - LeetCode 220 Solution

Q: How to solve Contains Duplicate III?

The provided code uses a bucket sort/pigeonhole principle-based approach in conjunction with a sliding window to efficiently solve the problem. It divides the number range into buckets of size `valueDiff + 1`. For each number in the array, we determine its bucket ID. Then, we check if: (1) the bucket is already occupied (meaning a duplicate within the allowed value difference exists), (2) either the adjacent buckets contain an element within the allowed value difference. The sliding window aspect is achieved by removing the oldest element (outside the indexDiff range) from the 'buckets' data structure as we iterate through the array.

Contains Duplicate III - Complete Solution Guide

Contains Duplicate III is LeetCode problem 220, a Hard 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 an integer array nums and two integers indexDiff and valueDiff . Find a pair of indices (i, j) such that: i != j , abs(i - j) <= indexDiff . abs(nums[i] - nums[j]) <= valueDiff , and Return true if such pair exists or false otherwise . Example 1: Input: nums = [1,2,3,1], indexDiff = 3, valueDiff = 0 Output: true Explanation: We can choose (i, j) = (0, 3). We satisfy the three conditions: i != j --> 0 != 3 abs(i - j) <= indexDiff --> abs(0 - 3) <= 3 abs(nums[i] - nums[j]) <= valueDi

Detailed Explanation

The problem requires us to find if there exist two distinct indices `i` and `j` in the input array `nums` such that the absolute difference between the indices is less than or equal to `indexDiff`, and the absolute difference between the values at those indices is less than or equal to `valueDiff`. In simpler terms, we need to find if there are two numbers in the array that are 'close' to each other both in terms of their value and their position in the array. The input consists of the array `nums`, the maximum allowed index difference `indexDiff`, and the maximum allowed value difference `valueDiff`. The output is a boolean value: `true` if such a pair exists, and `false` otherwise.

Solution Approach

The provided code uses a bucket sort/pigeonhole principle-based approach in conjunction with a sliding window to efficiently solve the problem. It divides the number range into buckets of size `valueDiff + 1`. For each number in the array, we determine its bucket ID. Then, we check if: (1) the bucket is already occupied (meaning a duplicate within the allowed value difference exists), (2) either the adjacent buckets contain an element within the allowed value difference. The sliding window aspect is achieved by removing the oldest element (outside the indexDiff range) from the 'buckets' data structure as we iterate through the array.

Step-by-Step Algorithm

Step 1: Handle edge cases. If `indexDiff` is non-positive or `valueDiff` is negative, return `false` as no valid pairs can exist.
Step 2: Initialize a hash map (buckets) to store the numbers and their corresponding bucket IDs.
Step 3: Calculate the bucket size as `valueDiff + 1`.
Step 4: Iterate through the array `nums` from left to right.
Step 5: For each number `num` at index `i`, calculate its bucket ID as `num // bucket_size`.
Step 6: Check if the current bucket ID already exists in the `buckets` hash map. If it does, return `true` (a duplicate within the allowed value difference is found).
Step 7: Check if the neighboring buckets (bucket_id - 1 and bucket_id + 1) exist in the `buckets` hash map. If they do, calculate the absolute difference between the current number `num` and the numbers stored in the neighboring buckets. If any of these differences are less than or equal to `valueDiff`, return `true`.
Step 8: Insert the current number `num` into the `buckets` hash map with its corresponding bucket ID.
Step 9: If the current index `i` is greater than or equal to `indexDiff`, remove the oldest number from the `buckets` hash map. The oldest number is at index `i - indexDiff` and its bucket is recalculated.
Step 10: If the loop completes without finding a suitable pair, return `false`.

Key Insights

Insight 1: Using a sliding window approach and checking all pairs within the window will result in O(n*k) time complexity (where 'k' is indexDiff), which can be too slow for large input sizes. The constraints indicate that an O(n) solution is expected.
Insight 2: The bucket sort/pigeonhole principle helps group nearby values into buckets, enabling efficient comparison of elements close in value. By using a bucket size of `valueDiff + 1`, any two elements within the same bucket or adjacent buckets must have a value difference less than or equal to `valueDiff`.
Insight 3: The sliding window is maintained by removing the element that falls out of the `indexDiff` range. This ensures that only elements within the acceptable index difference are considered.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(k)

Topics

This problem involves: Array, Sliding Window, Sorting, Bucket Sort, Ordered Set.

Companies

Asked at: Airbnb, Netflix, Palantir Technologies, Yandex.

Contains Duplicate III | LeetCode 220

Hard

Array Sliding Window Sorting Bucket Sort Ordered Set

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public boolean containsNearbyAlmostDuplicate(int[] nums, int indexDiff, int valueDiff) {
        if (indexDiff <= 0 || valueDiff < 0) {
            return false;
        }

        Map<Long, Integer> buckets = new HashMap<>();
        long bucketSize = (long)valueDiff + 1;

        for (int i = 0; i < nums.length; i++) {
            long num = (long) nums[i];
            long bucketId = num / bucketSize;

            if (buckets.containsKey(bucketId)) {
                return true;
            }

            if (buckets.containsKey(bucketId - 1) && Math.abs(num - (long)buckets.get(bucketId - 1)) <= valueDiff) {
                return true;
            }

            if (buckets.containsKey(bucketId + 1) && Math.abs(num - (long)buckets.get(bucketId + 1)) <= valueDiff) {
                return true;
            }

            buckets.put(bucketId, nums[i]);

            if (i >= indexDiff) {
                long oldNum = (long) nums[i - indexDiff];
                long oldBucketId = oldNum / bucketSize;
                if (buckets.containsKey(oldBucketId) && buckets.get(oldBucketId) == nums[i-indexDiff]) {
                    buckets.remove(oldBucketId);
                }
            }
        }

        return false;
    }
}

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Problem Info

Description

Problem 220: Contains Duplicate III - Detailed explanation and solution approaches.

Topics

Array Sliding Window Sorting Bucket Sort Ordered Set

Companies

AirbnbNetflixPalantir TechnologiesYandex

Stats

Difficulty:Hard

Languages:4

Companies:4

Problem #:220

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

10 steps

1
Step 1: Handle edge cases. If `indexDiff` is non-positive or `valueDiff` is negative, return `false` as no valid pairs can exist.
2
Step 2: Initialize a hash map (buckets) to store the numbers and their corresponding bucket IDs.
3
Step 3: Calculate the bucket size as `valueDiff + 1`.
4
Step 4: Iterate through the array `nums` from left to right.
5
Step 5: For each number `num` at index `i`, calculate its bucket ID as `num // bucket_size`.
6
Step 6: Check if the current bucket ID already exists in the `buckets` hash map. If it does, return `true` (a duplicate within the allowed value difference is found).
7
Step 7: Check if the neighboring buckets (bucket_id - 1 and bucket_id + 1) exist in the `buckets` hash map. If they do, calculate the absolute difference between the current number `num` and the numbers stored in the neighboring buckets. If any of these differences are less than or equal to `valueDiff`, return `true`.
8
Step 8: Insert the current number `num` into the `buckets` hash map with its corresponding bucket ID.
9
Step 9: If the current index `i` is greater than or equal to `indexDiff`, remove the oldest number from the `buckets` hash map. The oldest number is at index `i - indexDiff` and its bucket is recalculated.
10
Step 10: If the loop completes without finding a suitable pair, return `false`.

💡

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(k)

📝

Why?

The algorithm iterates through the array once, performing constant-time operations (hash table lookups, insertions, and deletions) within the loop. Therefore, the time complexity is directly proportional to the number of elements in the array, making it O(n). The space complexity is determined by the size of the `buckets` hash map. The hash map stores at most `indexDiff` key-value pairs because the sliding window removes elements outside of the allowed index range.

📖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

Iterate through all possible pairs of indices (i, j) in the array and check if the conditions `i != j`, `abs(i - j) <= indexDiff`, and `abs(nums[i] - nums[j]) <= valueDiff` are satisfied. This approach is simple to implement but inefficient.

Time: O(n^2), where n is the length of the array.Space: O(1)

Trade-offs: The brute force approach is simple but has a quadratic time complexity, making it unsuitable for large input sizes. It is only viable for very small input arrays where performance is not a concern. The provided solution with buckets should be preferred.

Sorted Set (TreeSet in Java / set in C++)

Use a sorted set (e.g., a self-balancing binary search tree) to maintain a sliding window of elements. For each element, search the sorted set for elements within the value difference. This is more efficient than brute force.

Time: O(n log k), where n is the length of the array and k is `indexDiff`.Space: O(k)

Trade-offs: The sorted set approach offers a better time complexity than brute force but is still less efficient than the bucket sort approach, which achieves O(n) time complexity. This approach has higher overhead due to log(k) factor.

✨

Key Insights

1Insight 1: Using a sliding window approach and checking all pairs within the window will result in O(n*k) time complexity (where 'k' is indexDiff), which can be too slow for large input sizes. The constraints indicate that an O(n) solution is expected.
2Insight 2: The bucket sort/pigeonhole principle helps group nearby values into buckets, enabling efficient comparison of elements close in value. By using a bucket size of `valueDiff + 1`, any two elements within the same bucket or adjacent buckets must have a value difference less than or equal to `valueDiff`.
3Insight 3: The sliding window is maintained by removing the element that falls out of the `indexDiff` range. This ensures that only elements within the acceptable index difference are considered.

📚

Key Terms Explained

For Students

Array

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

Sliding Window

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

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: `indexDiff` is 0 or negative. The solution correctly returns `false` because the problem requires `i != j` and `abs(i - j) <= indexDiff`. If `indexDiff` is 0 or negative, no such distinct indices can satisfy the condition.
Edge case 2: `valueDiff` is negative. The solution correctly returns `false` because absolute differences cannot be negative.
Edge case 3: Integer overflow. The solution casts integers to long long to avoid integer overflow when calculating differences and bucket IDs.
Edge case 4: Array contains duplicate numbers. The algorithm correctly handles cases where the array has multiple identical numbers and finds the duplicate numbers satisfying the condition.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling integer overflow when calculating the difference between numbers or the bucket ID. Using `long long` data type avoids this issue.

✗Common mistake 2: Incorrectly calculating the bucket size or bucket ID. The bucket size should be `valueDiff + 1` to ensure that elements in the same bucket or adjacent buckets are within the value difference.

✗Common mistake 3: Forgetting to remove the oldest element from the 'buckets' data structure to maintain the sliding window. This leads to considering elements outside the allowed index difference.

✗Common mistake 4: Not handling the edge cases of non-positive `indexDiff` or negative `valueDiff`. These cases should return `false` immediately.

✅

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: Fraud detection. Detecting fraudulent transactions that occur within a short time period and involve similar amounts can be done using this concept.

🏦Real-world scenario 2: Stock market analysis. Identifying stocks that have similar price fluctuations within a certain time window could be achieved by comparing the price changes with given tolerances (valueDiff, indexDiff).

🎮Real-world scenario 3: Data smoothing in sensor readings. Identifying and filtering out noisy sensor readings that deviate significantly from nearby readings within a specific time frame can use similar logic. This improves the accuracy and reliability of sensor data.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

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

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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 11, 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 13

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

@AlgoMasterJul 6

❓

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.

How do I determine the window size in sliding window problems?

For fixed-size windows (like max sum of k elements), window size is given. For variable-size windows (like smallest subarray with sum >= target), expand right pointer to meet condition, then shrink left pointer to find optimal. Track the best result during shrinking.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Contains Duplicate III"?

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

🤖 Need help with this problem?