What is the time complexity of Find the Maximum Length of a Good Subsequence II?

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

What is the space complexity of Find the Maximum Length of a Good Subsequence II?

The space complexity is O(n*k).

What data structures are used to solve Find the Maximum Length of a Good Subsequence II?

This problem uses Array, Hash Table, Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Find the Maximum Length of a Good Subsequence II?

Insight 1: Dynamic Programming is suitable as we are trying to find the maximum length by building up solutions from subproblems. Insight 2: We need to track the maximum length subsequence ending with a particular number and the number of differences encountered (up to `k`).

Which companies ask LeetCode 3177. Find the Maximum Length of a Good Subsequence II in interviews?

This problem is frequently asked at Snowflake.

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Find the Maximum Length of a Good Subsequence II - LeetCode 3177 Solution

Find the Maximum Length of a Good Subsequence II - Complete Solution Guide

Find the Maximum Length of a Good Subsequence II is LeetCode problem 3177, 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 a non-negative integer k . A sequence of integers seq is called good if there are at most k indices i in the range [0, seq.length - 2] such that seq[i] != seq[i + 1] . Return the maximum possible length of a good subsequence of nums . Example 1: Input: nums = [1,2,1,1,3], k = 2 Output: 4 Explanation: The maximum length subsequence is [ 1 , 2 , 1 , 1 ,3] . Example 2: Input: nums = [1,2,3,4,5,1], k = 0 Output: 2 Explanation: The maximum length subsequence is

Detailed Explanation

The problem requires us to find the maximum length of a "good" subsequence from a given integer array `nums`. A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements. A subsequence is considered "good" if it has at most `k` indices `i` where `seq[i] != seq[i+1]`. In other words, a good subsequence can have at most `k` adjacent pairs of different numbers. The input consists of an integer array `nums` and a non-negative integer `k`. The output is the maximum possible length of a good subsequence of `nums`. The constraints are: the length of `nums` is at most 5000, each element of `nums` is between 1 and 10^9, and `k` is between 0 and the minimum of 50 and the length of `nums`.

Solution Approach

The solution utilizes dynamic programming to find the maximum length of a good subsequence. It maintains a `dp` table (represented as a HashMap in Java and unordered_map in C++, and a 2D array in C) where `dp[num][j]` stores the maximum length of a good subsequence ending with `num` and having at most `j` differences. In addition to the DP table, the algorithm keeps track of the maximum (`max1`) and second maximum (`max2`) subsequence lengths for each value of `j` to efficiently compute the lengths of new subsequences. When processing a new number `num`, it updates the `dp` table by considering two options: extending an existing subsequence with the same last element or extending a subsequence with a different last element. The maximum and second maximum values are then updated for each `j` value.

Step-by-Step Algorithm

Step 1: Initialize the DP table `dp` to store the maximum length of subsequences ending with a particular number and number of differences.
Step 2: Initialize arrays `max1`, `val1`, and `max2` to track the maximum and second maximum subsequence lengths for each value of `j` (number of differences).
Step 3: Iterate through each number `num` in the input array `nums`.
Step 4: For each `num`, create a new array `new_lens_for_num` to store the updated subsequence lengths.
Step 5: Update `new_lens_for_num[0]` (the case when j = 0, i.e. no differences allowed) as 1 + `dp[num][0]`
Step 6: Iterate through `j` from 1 to `k`.
Step 7: Calculate `len_same`: the length obtained by extending a subsequence ending with `num` with the current number `num`.
Step 8: Calculate `len_diff`: the length obtained by extending a subsequence ending with a different number. This requires looking up the maximum subsequence length from the previous `j-1` differences, taking care to avoid extending with the same number again by using `max2` when necessary.
Step 9: Store the maximum of `len_same` and `len_diff` in `new_lens_for_num[j]`.
Step 10: Update `dp[num]` with `new_lens_for_num`.
Step 11: Update `max1`, `val1`, and `max2` based on the new subsequence lengths. If `new_lens_for_num[j]` is greater than the current `max1[j]`, update `max2[j]` with the old `max1[j]` (if the number is different), and set `max1[j]` to `new_lens_for_num[j]` and `val1[j]` to `num`. If `new_lens_for_num[j]` is greater than `max2[j]` but not `max1[j]` (and the number is different), update `max2[j]`.
Step 12: After iterating through all numbers in `nums`, return the maximum value in `max1`.

Key Insights

Insight 1: Dynamic Programming is suitable as we are trying to find the maximum length by building up solutions from subproblems.
Insight 2: We need to track the maximum length subsequence ending with a particular number and the number of differences encountered (up to `k`).
Insight 3: Maintaining both the maximum length and the second maximum length for each value of `k` allows efficient updating of DP states when the current number is equal to the previous maximum's number.

Complexity Analysis

Time Complexity: O(n*k)

Space Complexity: O(n*k)

Topics

This problem involves: Array, Hash Table, Dynamic Programming.

Companies

Asked at: Snowflake.

Find the Maximum Length of a Good Subsequence II | LeetCode 3177

Hard

Array Hash Table Dynamic Programming

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.Arrays;

class Solution {
    public int maximumLength(int[] nums, int k) {
        
        HashMap<Integer, int[]> dp = new HashMap<>();
        
        int[] max1 = new int[k + 1];
        Integer[] val1 = new Integer[k + 1];
        int[] max2 = new int[k + 1];
        
        for (int num : nums) {
            int[] new_lens_for_num = new int[k + 1];
            
            // Case j = 0
            int[] dp_num = dp.getOrDefault(num, new int[k + 1]);
            new_lens_for_num[0] = 1 + dp_num[0];
            
            // Case j > 0
            for (int j = 1; j <= k; j++) {
                // Option 1: Extend a subsequence with the same last element
                int len_same = 0;
                if (dp_num[j] > 0) {
                    len_same = 1 + dp_num[j];
                }
                
                // Option 2: Extend a subsequence with a different last element
                int len_diff = 0;
                int max_prev_len = (val1[j-1] != null && val1[j-1] != num) ? max1[j-1] : max2[j-1];
                if (max_prev_len > 0) {
                    len_diff = 1 + max_prev_len;
                }
                  
                new_lens_for_num[j] = Math.max(len_same, len_diff);
            }
            
            dp.put(num, new_lens_for_num);
            
            for (int j = 0; j <= k; j++) {
                int l = new_lens_for_num[j];
                if (l > max1[j]) {
                    if (val1[j] == null || val1[j] != num) {
                        max2[j] = max1[j];
                    }
                    max1[j] = l;
                    val1[j] = num;
                } else if (l > max2[j] && (val1[j] == null || val1[j] != num)) {
                    max2[j] = l;
                }
            }
        }
        
        int max_val = 0;
        for(int val : max1){
            max_val = Math.max(max_val, val);
        }
        
        return nums.length > 0 ? max_val : 0;
    }
}

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Problem Info

Description

Problem 3177: Find the Maximum Length of a Good Subsequence II - Detailed explanation and solution approaches.

Topics

Array Hash Table Dynamic Programming

Companies

Snowflake

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3177

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

12 steps

1
Step 1: Initialize the DP table `dp` to store the maximum length of subsequences ending with a particular number and number of differences.
2
Step 2: Initialize arrays `max1`, `val1`, and `max2` to track the maximum and second maximum subsequence lengths for each value of `j` (number of differences).
3
Step 3: Iterate through each number `num` in the input array `nums`.
4
Step 4: For each `num`, create a new array `new_lens_for_num` to store the updated subsequence lengths.
5
Step 5: Update `new_lens_for_num[0]` (the case when j = 0, i.e. no differences allowed) as 1 + `dp[num][0]`
6
Step 6: Iterate through `j` from 1 to `k`.
7
Step 7: Calculate `len_same`: the length obtained by extending a subsequence ending with `num` with the current number `num`.
8
Step 8: Calculate `len_diff`: the length obtained by extending a subsequence ending with a different number. This requires looking up the maximum subsequence length from the previous `j-1` differences, taking care to avoid extending with the same number again by using `max2` when necessary.
9
Step 9: Store the maximum of `len_same` and `len_diff` in `new_lens_for_num[j]`.
10
Step 10: Update `dp[num]` with `new_lens_for_num`.
11
Step 11: Update `max1`, `val1`, and `max2` based on the new subsequence lengths. If `new_lens_for_num[j]` is greater than the current `max1[j]`, update `max2[j]` with the old `max1[j]` (if the number is different), and set `max1[j]` to `new_lens_for_num[j]` and `val1[j]` to `num`. If `new_lens_for_num[j]` is greater than `max2[j]` but not `max1[j]` (and the number is different), update `max2[j]`.
12
Step 12: After iterating through all numbers in `nums`, return the maximum value in `max1`.

💡

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*k)

💾

Space Complexity

O(n*k)

📝

Why?

The time complexity is dominated by the nested loops. The outer loop runs `n` times, and the inner loop runs `k` times for each number. The space complexity comes from storing `dp` entries for each number and each value from 0 to k, plus the O(k) space for `max1`, `val1`, and `max2`.

📖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

Generate all possible subsequences, and then check if each subsequence is "good". Keep track of the maximum length of the "good" subsequences found.

Time: O(2^n * n), where n is the length of the `nums` array. Generating all subsequences takes O(2^n) time, and checking if each subsequence is good takes O(n) time.Space: O(n), where n is the length of the nums array

Trade-offs: Brute force is not feasible for large input sizes due to its exponential time complexity. Use it only for small `n`.

✨

Key Insights

1Insight 1: Dynamic Programming is suitable as we are trying to find the maximum length by building up solutions from subproblems.
2Insight 2: We need to track the maximum length subsequence ending with a particular number and the number of differences encountered (up to `k`).
3Insight 3: Maintaining both the maximum length and the second maximum length for each value of `k` allows efficient updating of DP states when the current number is equal to the previous maximum's number.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty input array `nums`. The solution returns 0 in this case, which is correct.
Edge case 2: `k` is 0. The algorithm correctly finds the maximum length of a subsequence where all adjacent elements are the same. This subsequence length will be no more than the frequency of the most frequent number.
Edge case 3: `k` is greater than or equal to `nums.length`. The algorithm will still work correctly as the inner loop iterates only up to `k`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where extending a subsequence with a different number requires looking at the maximum subsequence length from the previous 'j-1' differences, but the number is the same as the current number being processed. This is why maintaining both `max1` and `max2` is important.

✗Common mistake 2: Incorrectly updating `max1` and `max2` after processing each number. Ensuring that `max2` is updated only when a better length is found and the value is different from the value in `val1` is crucial.

✅

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: Data compression: Finding the longest subsequence with a limited number of changes can be useful in compressing data where you can allow a certain number of differences for better compression ratio.

🏦Real-world scenario 2: DNA sequencing: Identifying the longest segment of DNA that remains relatively unchanged over a specific region, allowing for a certain number of mutations or differences.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: September 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

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

@InterviewProJul 14

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

@CodeNinjaJul 7

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

@TechLeadJun 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.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Find the Maximum Length of a Good Subsequence II"?

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

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