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

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

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

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

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

Insight 1: The problem can be modeled as a Longest Increasing Subsequence (LIS) variant. The length of the strings and the Hamming distance constraint between adjacent strings acts as a condition for forming a valid subsequence. Insight 2: Dynamic Programming is a suitable approach. We can build a DP array to store the length of the longest valid subsequence ending at each index.

Which companies ask LeetCode 2901. Longest Unequal Adjacent Groups Subsequence II in interviews?

This problem is frequently asked at fourkites.

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Longest Unequal Adjacent Groups Subsequence II - LeetCode 2901 Solution

Longest Unequal Adjacent Groups Subsequence II - Complete Solution Guide

Longest Unequal Adjacent Groups Subsequence II is LeetCode problem 2901, a Medium 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 an array groups , both arrays having length n . The hamming distance between two strings of equal length is the number of positions at which the corresponding characters are different . You need to select the longest subsequence from an array of indices [0, 1, ..., n - 1] , such that for the subsequence denoted as [i 0 , i 1 , ..., i k-1 ] having length k , the following holds: For adjacent indices in the subsequence, their corresponding groups are unequa

Detailed Explanation

The problem asks us to find the longest subsequence from a given array of words and their corresponding groups, satisfying two conditions: (1) Adjacent words in the subsequence must belong to different groups. (2) Adjacent words must have the same length and a Hamming distance of exactly 1 (differ by only one character). We need to return the subsequence of words that satisfies these conditions and has the maximum possible length. Note that the input words are distinct, but the groups may contain duplicates.

Solution Approach

The solution uses dynamic programming to find the length of the longest subsequence that satisfies the given conditions. It iterates through each word in the input array. For each word, it checks all preceding words to see if they can extend the subsequence while adhering to both the group difference and Hamming distance requirements. It keeps track of the length of the longest subsequence ending at each index and also maintains a 'parent' array to reconstruct the subsequence after calculating the DP array. Finally, the code reconstructs the longest valid subsequence by backtracking from the index with the maximum DP value using the parent array.

Step-by-Step Algorithm

Step 1: Initialize two arrays: `dp` of size `n` (where `n` is the number of words) to store the length of the longest subsequence ending at each index, and `parent` of size `n` to keep track of the preceding index in the longest subsequence.
Step 2: Iterate through the `words` array from `i = 0` to `n-1`. Initialize `dp[i] = 1` and `parent[i] = -1` (meaning the subsequence initially contains only the current word).
Step 3: For each word at index `i`, iterate through all preceding words from `j = 0` to `i-1`.
Step 4: Inside the inner loop, check two conditions: (a) `groups[j] != groups[i]` (the groups must be different), and (b) `len(words[j]) == len(words[i])` (the lengths must be equal).
Step 5: If both conditions in step 4 are true, calculate the Hamming distance between `words[i]` and `words[j]`.
Step 6: If the Hamming distance is exactly 1, check if extending the subsequence ending at index `j` by adding `words[i]` results in a longer subsequence than the current length of the subsequence ending at index `i` (i.e., `1 + dp[j] > dp[i]`).
Step 7: If the condition in step 6 is true, update `dp[i] = 1 + dp[j]` and `parent[i] = j`.
Step 8: After the inner loops complete, find the index `end_index` with the maximum value in the `dp` array. This index marks the end of the longest subsequence.
Step 9: Reconstruct the longest subsequence by backtracking from `end_index` using the `parent` array. Start from `curr = end_index`, add `words[curr]` to the result, and then set `curr = parent[curr]` until `curr` becomes -1.
Step 10: Reverse the resulting list of words (as we built it in reverse order) and return it.

Key Insights

Insight 1: The problem can be modeled as a Longest Increasing Subsequence (LIS) variant. The length of the strings and the Hamming distance constraint between adjacent strings acts as a condition for forming a valid subsequence.
Insight 2: Dynamic Programming is a suitable approach. We can build a DP array to store the length of the longest valid subsequence ending at each index.
Insight 3: The group constraint and Hamming distance constraint are crucial and need to be checked before extending the subsequence.

Complexity Analysis

Time Complexity: O(n^2*m)

Space Complexity: O(n)

Topics

This problem involves: Array, String, Dynamic Programming.

Companies

Asked at: fourkites.

Longest Unequal Adjacent Groups Subsequence II | LeetCode 2901

Medium

Array String Dynamic Programming

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

Solution Code

Available in:

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

class Solution {
    public List<String> getWordsInLongestSubsequence(String[] words, int[] groups) {
        int n = words.length;
        
        int[] dp = new int[n];
        int[] parent = new int[n];
        
        for (int i = 0; i < n; i++) {
            dp[i] = 1;
            parent[i] = -1;
        }
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (groups[j] != groups[i] && words[j].length() == words[i].length()) {
                    int diff_count = 0;
                    for (int k = 0; k < words[i].length(); k++) {
                        if (words[i].charAt(k) != words[j].charAt(k)) {
                            diff_count++;
                        }
                        if (diff_count > 1) {
                            break;
                        }
                    }
                    
                    if (diff_count == 1) {
                        if (1 + dp[j] > dp[i]) {
                            dp[i] = 1 + dp[j];
                            parent[i] = j;
                        }
                    }
                }
            }
        }
        
        int max_len = 0;
        int end_index = 0;
        for (int i = 0; i < n; i++) {
            if (dp[i] > max_len) {
                max_len = dp[i];
                end_index = i;
            }
        }
        
        List<String> res = new ArrayList<>();
        int curr = end_index;
        while (curr != -1) {
            res.add(words[curr]);
            curr = parent[curr];
        }
        
        List<String> reversedRes = new ArrayList<>();
        for (int i = res.size() - 1; i >= 0; i--) {
            reversedRes.add(res.get(i));
        }

        return reversedRes;
    }
}

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Problem Info

Description

Problem 2901: Longest Unequal Adjacent Groups Subsequence II - Detailed explanation and solution approaches.

Topics

Array String Dynamic Programming

Companies

fourkites

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2901

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 arrays: `dp` of size `n` (where `n` is the number of words) to store the length of the longest subsequence ending at each index, and `parent` of size `n` to keep track of the preceding index in the longest subsequence.
2
Step 2: Iterate through the `words` array from `i = 0` to `n-1`. Initialize `dp[i] = 1` and `parent[i] = -1` (meaning the subsequence initially contains only the current word).
3
Step 3: For each word at index `i`, iterate through all preceding words from `j = 0` to `i-1`.
4
Step 4: Inside the inner loop, check two conditions: (a) `groups[j] != groups[i]` (the groups must be different), and (b) `len(words[j]) == len(words[i])` (the lengths must be equal).
5
Step 5: If both conditions in step 4 are true, calculate the Hamming distance between `words[i]` and `words[j]`.
6
Step 6: If the Hamming distance is exactly 1, check if extending the subsequence ending at index `j` by adding `words[i]` results in a longer subsequence than the current length of the subsequence ending at index `i` (i.e., `1 + dp[j] > dp[i]`).
7
Step 7: If the condition in step 6 is true, update `dp[i] = 1 + dp[j]` and `parent[i] = j`.
8
Step 8: After the inner loops complete, find the index `end_index` with the maximum value in the `dp` array. This index marks the end of the longest subsequence.
9
Step 9: Reconstruct the longest subsequence by backtracking from `end_index` using the `parent` array. Start from `curr = end_index`, add `words[curr]` to the result, and then set `curr = parent[curr]` until `curr` becomes -1.
10
Step 10: Reverse the resulting list of words (as we built it in reverse order) and return it.

💡

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*m)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is derived from the nested loops checking all possible pairs of words, and the Hamming distance calculation for each pair. The space complexity comes primarily from the DP and parent arrays used to store intermediate results during the dynamic programming process.

📖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

Top-Down Dynamic Programming with Memoization

Instead of the bottom-up DP approach, we can use a top-down approach with memoization. This involves defining a recursive function that explores all possible subsequences and memoizes the results to avoid redundant computations.

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

Trade-offs: Top-down DP can sometimes be more intuitive to implement, but it might have a higher overhead due to recursion. The bottom-up approach is usually more efficient in terms of constant factors.

✨

Key Insights

1Insight 1: The problem can be modeled as a Longest Increasing Subsequence (LIS) variant. The length of the strings and the Hamming distance constraint between adjacent strings acts as a condition for forming a valid subsequence.
2Insight 2: Dynamic Programming is a suitable approach. We can build a DP array to store the length of the longest valid subsequence ending at each index.
3Insight 3: The group constraint and Hamming distance constraint are crucial and need to be checked before extending the subsequence.

📚

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.

String

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

Edge Cases

Edge case 1: Empty input arrays. The code implicitly handles this by not entering the loops, and returning an empty list.
Edge case 2: No valid subsequence exists. In this case, the `dp` array will contain only 1s, and the algorithm will return the longest subsequence of length 1, which consists of a single word.
Edge case 3: All words have different lengths or belong to the same group. The algorithm will correctly find the longest subsequence of length 1 in such cases.
Edge case 4: The input array contains only one word. The algorithm will return a list containing only that word, as it will be considered a subsequence of length 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check both the group difference and Hamming distance constraints. Failing to check one of these conditions will lead to incorrect subsequences.

✗Common mistake 2: Not reversing the resulting subsequence after backtracking. The subsequence is built in reverse order during backtracking, so it needs to be reversed before being returned.

✗Common mistake 3: Incorrectly calculating the Hamming distance. Ensure the Hamming distance calculation only increments the count for different characters at corresponding positions.

✅

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: Bioinformatics - Identifying sequences of DNA or protein that are similar but belong to different functional groups. The Hamming distance can represent the number of mutations between sequences.

🏦Real-world scenario 2: Recommendation Systems - Finding items that are closely related but belong to different categories to diversify recommendations. For example, recommending movies from different genres but with actors who have similar roles.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: June 22, 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

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

@AlgoMasterJul 14

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

@InterviewProJul 7

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

@CodeNinjaJun 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 is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 Unequal Adjacent Groups Subsequence II"?

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

🤖 Need help with this problem?