What is the time complexity of Longest Unequal Adjacent Groups Subsequence I?

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

What data structures are used to solve Longest Unequal Adjacent Groups Subsequence I?

This problem uses Array, String, Dynamic Programming, Greedy. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Longest Unequal Adjacent Groups Subsequence I?

Insight 1: The problem can be solved greedily. There's no need to explore all possible subsequences; we can build the longest alternating subsequence iteratively. Insight 2: A simple iteration through the `words` and `groups` arrays is sufficient. We only need to keep track of the group value of the last selected string.

Which companies ask LeetCode 2900. Longest Unequal Adjacent Groups Subsequence I in interviews?

This problem is frequently asked at ZS Associates, fourkites.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Longest Unequal Adjacent Groups Subsequence I - LeetCode 2900 Solution

Longest Unequal Adjacent Groups Subsequence I - Complete Solution Guide

Longest Unequal Adjacent Groups Subsequence I is LeetCode problem 2900, 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 a string array words and a binary array groups both of length n . A subsequence of words is alternating if for any two consecutive strings in the sequence, their corresponding elements at the same indices in groups are different (that is, there cannot be consecutive 0 or 1). Your task is to select the longest alternating subsequence from words . Return the selected subsequence. If there are multiple answers, return any of them. Note: The elements in words are distinct. Example 1: I

Detailed Explanation

The problem asks you to find the longest subsequence from a given array of strings (`words`) such that consecutive strings in the subsequence have different corresponding group values in a separate binary array (`groups`). In simpler terms, if you pick a string from `words`, the next string you pick must have a different group value (0 or 1) at its corresponding index in `groups`. The output is the longest subsequence of strings that satisfies this alternating group condition. The input `words` array contains distinct strings, and `groups` contains only 0s and 1s.

Solution Approach

The provided solution uses a greedy approach. It initializes the result with the first word and its group value. Then, it iterates through the remaining words. If the current word's group value is different from the previously selected word's group value, it adds the current word to the result. This continues until all words are processed. The final result is the longest alternating subsequence.

Step-by-Step Algorithm

Step 1: Initialize the result list with the first word from `words`.
Step 2: Initialize a variable `prev_group` to store the group value of the first word (from `groups`).
Step 3: Iterate through the remaining words and their corresponding group values, starting from the second word.
Step 4: If the current word's group value is different from `prev_group`, append the current word to the result list and update `prev_group`.
Step 5: Return the result list containing the longest alternating subsequence.

Key Insights

Insight 1: The problem can be solved greedily. There's no need to explore all possible subsequences; we can build the longest alternating subsequence iteratively.
Insight 2: A simple iteration through the `words` and `groups` arrays is sufficient. We only need to keep track of the group value of the last selected string.
Insight 3: Handling the base case where `words` is empty or has only one element is necessary to avoid errors.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, String, Dynamic Programming, Greedy.

Companies

Asked at: ZS Associates, fourkites.

Longest Unequal Adjacent Groups Subsequence I | LeetCode 2900

Easy

Array String Dynamic Programming Greedy

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<String> getLongestSubsequence(String[] words, int[] groups) {
        List<String> result = new ArrayList<>();
        if (words == null || words.length == 0) {
            return result;
        }

        result.add(words[0]);
        int prevGroup = groups[0];

        for (int i = 1; i < words.length; i++) {
            if (groups[i] != prevGroup) {
                result.add(words[i]);
                prevGroup = groups[i];
            }
        }

        return result;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2900: Longest Unequal Adjacent Groups Subsequence I - Detailed explanation and solution approaches.

Topics

Array String Dynamic Programming Greedy

Companies

ZS Associatesfourkites

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:2900

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 the result list with the first word from `words`.
2
Step 2: Initialize a variable `prev_group` to store the group value of the first word (from `groups`).
3
Step 3: Iterate through the remaining words and their corresponding group values, starting from the second word.
4
Step 4: If the current word's group value is different from `prev_group`, append the current word to the result list and update `prev_group`.
5
Step 5: Return the result list containing the longest alternating subsequence.

💡

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 `words` and `groups` arrays once, which takes O(n) time, where n is the length of the arrays. The space complexity is O(n) because, in the worst case, the resulting subsequence could contain all the words (if they alternate perfectly), requiring O(n) space to store.

📖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

Recursive Approach

A recursive approach could explore all possible subsequences and return the longest one that satisfies the alternating condition. This would involve backtracking.

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

Trade-offs: This approach is significantly less efficient than the greedy approach due to its exponential time complexity. The greedy approach is preferred for its efficiency in this case.

✨

Key Insights

1Insight 1: The problem can be solved greedily. There's no need to explore all possible subsequences; we can build the longest alternating subsequence iteratively.
2Insight 2: A simple iteration through the `words` and `groups` arrays is sufficient. We only need to keep track of the group value of the last selected string.
3Insight 3: Handling the base case where `words` is empty or has only one element is necessary to avoid errors.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty `words` array. The solution correctly handles this by returning an empty list.
Edge case 2: `words` array with only one element. The solution correctly handles this by returning a list containing only that element.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle edge cases (empty or single-element input). This can lead to runtime errors or incorrect results.

✗Common mistake 2: Incorrectly comparing group values. Failing to ensure that consecutive elements in the subsequence have different group values will result in an invalid subsequence.

✅

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: Selecting tasks from a list where each task has a priority level (0 or 1), ensuring that you alternate between high (1) and low (0) priority tasks to maintain balance.

🏦Real-world scenario 2: Optimizing a network connection by alternating between two different communication channels or protocols to improve reliability or throughput.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ 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: 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.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 Unequal Adjacent Groups Subsequence I"?

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

🤖 Need help with this problem?