What is the time complexity of Longest Strictly Increasing or Strictly Decreasing Subarray?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Longest Strictly Increasing or Strictly Decreasing Subarray?

This problem uses Array. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Longest Strictly Increasing or Strictly Decreasing Subarray?

Insight 1: Iterate through the array, checking for both increasing and decreasing subsequences starting at each index. Insight 2: Maintain a `maxLength` variable to track the longest subsequence encountered so far.

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Longest Strictly Increasing or Strictly Decreasing Subarray - LeetCode 3105 Solution

Q: Which companies ask LeetCode 3105. Longest Strictly Increasing or Strictly Decreasing Subarray in interviews?

This problem is frequently asked at Larsen & Toubro.

Longest Strictly Increasing or Strictly Decreasing Subarray - Complete Solution Guide

Longest Strictly Increasing or Strictly Decreasing Subarray is LeetCode problem 3105, a Easy 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 array of integers nums . Return the length of the longest subarray of nums which is either strictly increasing or strictly decreasing . Example 1: Input: nums = [1,4,3,3,2] Output: 2 Explanation: The strictly increasing subarrays of nums are [1] , [2] , [3] , [3] , [4] , and [1,4] . The strictly decreasing subarrays of nums are [1] , [2] , [3] , [3] , [4] , [3,2] , and [4,3] . Hence, we return 2 . Example 2: Input: nums = [3,3,3,3] Output: 1 Explanation: The strictly increasing

Detailed Explanation

The problem asks you to find the length of the longest subarray within a given array `nums` that is either strictly increasing or strictly decreasing. A strictly increasing subarray means each element is greater than the previous one, and a strictly decreasing subarray means each element is smaller than the previous one. The input is an array of integers, and the output is a single integer representing the length of the longest such subarray.

Solution Approach

The provided solution uses a brute-force approach. It iterates through each element of the input array. For each element, it checks the length of the longest strictly increasing and strictly decreasing subsequences starting from that element. It updates the `maxLength` variable accordingly and returns the final `maxLength` at the end. This approach considers all possible subarrays starting at each index.

Step-by-Step Algorithm

Step 1: Initialize `maxLength` to 1 (handling the case of a single-element array).
Step 2: Iterate through the `nums` array using a loop (outer loop).
Step 3: For each element, start an inner loop to check for increasing subsequences. Count the consecutive increasing elements. Update `maxLength` if a longer increasing subsequence is found.
Step 4: Similarly, for each element, start another inner loop to check for decreasing subsequences. Count the consecutive decreasing elements. Update `maxLength` if a longer decreasing subsequence is found.
Step 5: After iterating through all elements, return `maxLength`.

Key Insights

Insight 1: Iterate through the array, checking for both increasing and decreasing subsequences starting at each index.
Insight 2: Maintain a `maxLength` variable to track the longest subsequence encountered so far.
Insight 3: Handle edge cases such as empty arrays, arrays with one element, and arrays with only constant values.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Array.

Companies

Asked at: Larsen & Toubro.

Longest Strictly Increasing or Strictly Decreasing Subarray | LeetCode 3105

Easy

Array

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public int longestMonotonicSubarray(int[] nums) {
        int n = nums.length;
        if (n <= 1) {
            return n;
        }
        int maxLength = 1;
        for (int i = 0; i < n; i++) {
            int increasingLength = 1;
            for (int j = i + 1; j < n; j++) {
                if (nums[j] > nums[j - 1]) {
                    increasingLength++;
                } else {
                    break;
                }
            }
            maxLength = Math.max(maxLength, increasingLength);

            int decreasingLength = 1;
            for (int j = i + 1; j < n; j++) {
                if (nums[j] < nums[j - 1]) {
                    decreasingLength++;
                } else {
                    break;
                }
            }
            maxLength = Math.max(maxLength, decreasingLength);
        }
        return maxLength;
    }
}

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Problem Info

Description

Problem 3105: Longest Strictly Increasing or Strictly Decreasing Subarray - Detailed explanation and solution approaches.

Topics

Array

Companies

Larsen & Toubro

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3105

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

5 steps

1
Step 1: Initialize `maxLength` to 1 (handling the case of a single-element array).
2
Step 2: Iterate through the `nums` array using a loop (outer loop).
3
Step 3: For each element, start an inner loop to check for increasing subsequences. Count the consecutive increasing elements. Update `maxLength` if a longer increasing subsequence is found.
4
Step 4: Similarly, for each element, start another inner loop to check for decreasing subsequences. Count the consecutive decreasing elements. Update `maxLength` if a longer decreasing subsequence is found.
5
Step 5: After iterating through all elements, return `maxLength`.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The nested loops dominate the time complexity. The outer loop iterates `n` times, and the inner loops also iterate up to `n` times in the worst case (e.g., a strictly increasing or decreasing array). Therefore, the time complexity is O(n^2). The algorithm uses a constant amount of extra space to store variables like `maxLength`, `curr_len`, `increasingLength`, `decreasingLength`, etc., regardless of the input size, leading to O(1) space complexity.

📖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

Dynamic Programming

While not strictly necessary given the constraints (n <= 50), a dynamic programming approach could be used to slightly optimize the solution. Two arrays could track the lengths of increasing and decreasing subsequences ending at each index.

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

Trade-offs: Dynamic programming offers a linear time solution but at the cost of linear space complexity. For this specific problem with small constraints, the brute force method is sufficiently efficient.

✨

Key Insights

1Insight 1: Iterate through the array, checking for both increasing and decreasing subsequences starting at each index.
2Insight 2: Maintain a `maxLength` variable to track the longest subsequence encountered so far.
3Insight 3: Handle edge cases such as empty arrays, arrays with one element, and arrays with only constant values.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: Empty array. The code handles this by checking `numsSize` or `n` before proceeding. If the array is empty, it returns 0.
Edge case 2: Array with a single element. The code returns 1 in this case, as the subarray length is 1.
Edge case 3: Array with all the same elements. The code correctly identifies that the length of the longest monotonic subarray is 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check both increasing and decreasing subsequences. The solution must explicitly check both types to satisfy the problem requirements.

✗Common mistake 2: Incorrectly handling the base cases (empty or single-element arrays). The code needs to return 0 for an empty array and 1 for a single-element array.

✅

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: Analyzing stock prices to find the longest period of continuous increase or decrease. This could be useful for identifying trends in the market.

🏦Real-world scenario 2: Detecting patterns in sensor data. Identifying long periods of consistent increase or decrease in temperature, pressure, or other sensor readings can help diagnose equipment issues or environmental changes.

💡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.

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

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 16, 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

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

@InterviewProJun 30

❓

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.

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 "Longest Strictly Increasing or Strictly Decreasing Subarray"?

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

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