What is the time complexity of Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit?

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

What data structures are used to solve Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit?

This problem uses Array, Queue, Sliding Window, Monotonic Queue. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit?

Insight 1: Brute-force enumeration of all subarrays is inefficient due to its high time complexity (O(n^2) or O(n^3) depending on how max absolute difference is calculated). Insight 2: We can use a sliding window approach to efficiently explore subarrays. The challenge is maintaining the maximum and minimum values within the current window.

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Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit - LeetCode 1438 Solution

Q: Which companies ask LeetCode 1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit in interviews?

This problem is frequently asked at Capital One, Databricks, Moloco, Nutanix, PhonePe and 2 more companies.

Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit - Complete Solution Guide

Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit is LeetCode problem 1438, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3.

Problem Statement

Given an array of integers nums and an integer limit , return the size of the longest non-empty subarray such that the absolute difference between any two elements of this subarray is less than or equal to limit . Example 1: Input: nums = [8,2,4,7], limit = 4 Output: 2 Explanation: All subarrays are: [8] with maximum absolute diff |8-8| = 0 <= 4. [8,2] with maximum absolute diff |8-2| = 6 > 4. [8,2,4] with maximum absolute diff |8-2| = 6 > 4. [8,2,4,7] with maximum absolute diff |8-2| = 6 > 4. [

Detailed Explanation

The problem asks us to find the length of the longest continuous subarray within a given array `nums` such that the absolute difference between any two elements in that subarray is less than or equal to a given `limit`. We need to iterate through all possible subarrays, calculate the maximum absolute difference within each, and return the length of the longest subarray that satisfies the limit constraint. The input is an array of integers `nums` and an integer `limit`. The output is an integer representing the length of the longest valid subarray.

Solution Approach

The solution utilizes a sliding window technique along with two monotonic deques: one for maintaining the maximum value within the window (max_deque) and another for maintaining the minimum value (min_deque). The window expands by incrementing the 'right' pointer. When the absolute difference between the maximum and minimum values in the window exceeds the given 'limit', the window shrinks by incrementing the 'left' pointer. The deques are updated to maintain their monotonic properties (decreasing for max_deque, increasing for min_deque) as new elements are added and old elements are removed. The maximum length of a valid subarray is tracked throughout the process.

Step-by-Step Algorithm

Step 1: Initialize two double-ended queues (deques), `max_deque` and `min_deque`, to store indices of potential maximum and minimum elements within the current window.
Step 2: Initialize `left` to 0, representing the left boundary of the sliding window.
Step 3: Initialize `max_len` to 0, representing the maximum length of a valid subarray found so far.
Step 4: Iterate through the `nums` array with the `right` pointer, expanding the sliding window.
Step 5: For each `right`, update `max_deque`: remove elements from the back of `max_deque` that are smaller than or equal to `nums[right]` to maintain a decreasing order. Append `right` to `max_deque`.
Step 6: Similarly, update `min_deque`: remove elements from the back of `min_deque` that are greater than or equal to `nums[right]` to maintain an increasing order. Append `right` to `min_deque`.
Step 7: Check if the absolute difference between the maximum and minimum elements within the window (i.e., `nums[max_deque[0]] - nums[min_deque[0]]`) is greater than `limit`.
Step 8: If the difference exceeds the `limit`, shrink the window by incrementing `left`. Remove `left` from the front of `max_deque` and `min_deque` if it is the index at the front of the queue. Repeat until the difference is within the limit.
Step 9: Update `max_len` with the maximum of its current value and the current window size (`right - left + 1`).
Step 10: After iterating through the entire `nums` array, return `max_len`.

Key Insights

Insight 1: Brute-force enumeration of all subarrays is inefficient due to its high time complexity (O(n^2) or O(n^3) depending on how max absolute difference is calculated).
Insight 2: We can use a sliding window approach to efficiently explore subarrays. The challenge is maintaining the maximum and minimum values within the current window.
Insight 3: Monotonic queues (deques) are excellent for keeping track of the maximum and minimum elements in a sliding window in O(1) amortized time per element.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Queue, Sliding Window, Monotonic Queue.

Companies

Asked at: Capital One, Databricks, Moloco, Nutanix, PhonePe, Visa, eBay.

Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit | LeetCode 1438

Medium

Array Queue Sliding Window Monotonic Queue

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

Solution Code

Available in:

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Problem Info

Description

Problem 1438: Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit - Detailed explanation and solution approaches.

Topics

Array Queue Sliding Window Monotonic Queue

Companies

Capital OneDatabricksMolocoNutanixPhonePeVisaeBay

Stats

Difficulty:Medium

Languages:2

Companies:7

Problem #:1438

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: Initialize two double-ended queues (deques), `max_deque` and `min_deque`, to store indices of potential maximum and minimum elements within the current window.
2
Step 2: Initialize `left` to 0, representing the left boundary of the sliding window.
3
Step 3: Initialize `max_len` to 0, representing the maximum length of a valid subarray found so far.
4
Step 4: Iterate through the `nums` array with the `right` pointer, expanding the sliding window.
5
Step 5: For each `right`, update `max_deque`: remove elements from the back of `max_deque` that are smaller than or equal to `nums[right]` to maintain a decreasing order. Append `right` to `max_deque`.
6
Step 6: Similarly, update `min_deque`: remove elements from the back of `min_deque` that are greater than or equal to `nums[right]` to maintain an increasing order. Append `right` to `min_deque`.
7
Step 7: Check if the absolute difference between the maximum and minimum elements within the window (i.e., `nums[max_deque[0]] - nums[min_deque[0]]`) is greater than `limit`.
8
Step 8: If the difference exceeds the `limit`, shrink the window by incrementing `left`. Remove `left` from the front of `max_deque` and `min_deque` if it is the index at the front of the queue. Repeat until the difference is within the limit.
9
Step 9: Update `max_len` with the maximum of its current value and the current window size (`right - left + 1`).
10
Step 10: After iterating through the entire `nums` array, return `max_len`.

💡

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 each element in `nums` is visited at most a constant number of times (added to and removed from the deques). The `left` pointer also only increments up to `n`. The space complexity is O(n) because in the worst case, the deques could store all the indices of the `nums` array (e.g., when the array is sorted in increasing or decreasing order, and `limit` is large).

📖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

Priority Queue (Heap)

Use two priority queues (one min-heap and one max-heap) to track the minimum and maximum values in the window. However, removing elements from the middle of a heap is inefficient (O(n)), so a lazy deletion approach is needed.

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

Trade-offs: The heap approach has a higher time complexity due to the logarithmic operations of the priority queues. The Monotonic Queue approach is generally more efficient.

Segment Tree / Range Minimum/Maximum Query

Using a segment tree would allow you to quickly find the max and min of any window of the array.

Time: O(n log n) for construction, O(log n) for each query, leading to O(n log n) overallSpace: O(n)

Trade-offs: While queries can be made in O(log n) time, building the segment tree requires O(n log n). It's overkill for this particular problem, since it requires more complex setup than the Monotonic Queue approach. This is best if the array `nums` is fixed and many queries are needed.

✨

Key Insights

1Insight 1: Brute-force enumeration of all subarrays is inefficient due to its high time complexity (O(n^2) or O(n^3) depending on how max absolute difference is calculated).
2Insight 2: We can use a sliding window approach to efficiently explore subarrays. The challenge is maintaining the maximum and minimum values within the current window.
3Insight 3: Monotonic queues (deques) are excellent for keeping track of the maximum and minimum elements in a sliding window in O(1) amortized time per element.

📚

Key Terms Explained

For Students

Array

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

Queue

FIFO structure - First In, First Out. Like a line at a store!

Sliding Window

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

Edge Cases

Edge case 1: Empty input array: The code implicitly handles this case because the loop won't execute, and the initial `max_len` value of 0 will be returned.
Edge case 2: `limit` is 0: The code correctly identifies subarrays where all elements are equal. The deques will only contain indices of equal elements until a different element forces the window to shrink.
Edge case 3: `limit` is very large: The code will find the longest possible subarray without any constraints, which is the entire array if the `limit` is greater than or equal to the maximum absolute difference in the input array.
Edge case 4: All elements are the same: The deques will only contain indices of these elements and the window will grow as large as possible until the end of the input array.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to maintain the monotonic properties of the deques. The deques must be strictly decreasing (max_deque) or strictly increasing (min_deque) to ensure that the element at the front is always the maximum or minimum within the current window.

✗Common mistake 2: Incorrectly updating the `left` pointer. When shrinking the window, it's crucial to only increment `left` and remove from the deques until the absolute difference is within the limit. Also, remember to remove the index `left` from the front of the deque *only if* it is the index at the front of the deque.

✗Common mistake 3: Not considering all possible subarrays. A sliding window needs to ensure that all valid window sizes are explored.

✅

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: Stock price analysis: Finding the longest period where the price fluctuation (difference between the highest and lowest price) is within a certain limit.

🏦Real-world scenario 2: Signal processing: Identifying the longest segment of a signal where the amplitude remains within a specified range.

🎮Real-world scenario 3: Network traffic analysis: Determining the longest interval where the network latency stays below a certain threshold.

💡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

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

⚠️ 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: November 26, 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

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

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 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.

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 "Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit"?

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

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