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

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

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

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

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

Insight 1: Dynamic Programming: The problem's optimal substructure makes dynamic programming a suitable approach. We can build up the solution by considering smaller subproblems. Insight 2: State Representation: A key challenge is representing the state in the DP. We need to track the subsequence length and the number of allowed differences (k).

Which companies ask LeetCode 3176. Find the Maximum Length of a Good Subsequence I in interviews?

This problem is frequently asked at Snowflake.

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Find the Maximum Length of a Good Subsequence I - LeetCode 3176 Solution

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

Find the Maximum Length of a Good Subsequence I is LeetCode problem 3176, 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 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 asks us to find the maximum length of a "good" subsequence within a given integer array `nums`, where a subsequence is "good" if it has at most `k` indices `i` such that `seq[i] != seq[i+1]`. In simpler terms, we want to find the longest possible subsequence where there are at most `k` differences between consecutive elements. The inputs are the array `nums` and the integer `k`. The output is the maximum length of a good subsequence.

Solution Approach

The provided code uses dynamic programming to solve the problem. `dp[j][num]` stores the maximum length of a good subsequence ending with `num`, having used `j` allowed differences. The `max1` and `max2` arrays store the largest and second largest lengths ending with different values for each possible number of differences allowed from 0 to `k`. This optimization reduces the time complexity of iterating through all possible last elements to find the best length.

Step-by-Step Algorithm

Step 1: Initialize DP and Max Length Arrays: Create a 2D DP array `dp` where `dp[j][num]` represents the maximum length of a good subsequence ending with `num` and using at most `j` differences. Also initialize two arrays `max1` and `max2` to store the two maximum lengths of the subsequences we have seen so far, and their respective ending element values. Note that each row of 'dp' is a dictionary.
Step 2: Iterate Through the Input Array: Iterate through the input array `nums`, element by element.
Step 3: Calculate Lengths with Same and Different Numbers: For each `num` in `nums` and for each allowed difference count `j` (from 0 to `k`), calculate two potential lengths: `len_same` (length of subsequence if the current `num` is the same as the last element of the subsequence) and `len_diff` (length if the current `num` is different).
Step 4: Update Lengths: Determine the new max length given the current num. Store it in 'new_lens_for_num'.
Step 5: Update DP and Max Length Arrays: Update the dp table and the max arrays using the new max length.
Step 6: Find Maximum Length: After iterating through the entire input array, find the maximum length subsequence possible by checking the maximum values stored in the max length array.

Key Insights

Insight 1: Dynamic Programming: The problem's optimal substructure makes dynamic programming a suitable approach. We can build up the solution by considering smaller subproblems.
Insight 2: State Representation: A key challenge is representing the state in the DP. We need to track the subsequence length and the number of allowed differences (k).
Insight 3: Optimization: The original approach keeps track of the maximum length good subsequence. Because we only need to find the maximum length subsequence, the 'max1' and 'max2' arrays are used to optimize the DP process.

Complexity Analysis

Time Complexity: O(nk)

Space Complexity: O(nk)

Topics

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

Companies

Asked at: Snowflake.

Find the Maximum Length of a Good Subsequence I | LeetCode 3176

Medium

Array Hash Table Dynamic Programming

Time: O(nk)Space: O(nk)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public int maximumLength(int[] nums, int k) {
        int n = nums.length;
        if (n == 0) {
            return 0;
        }

        List<Map<Integer, Integer>> dp = new ArrayList<>();
        for (int i = 0; i <= k; i++) {
            dp.add(new HashMap<>());
        }

        List<Pair> max1 = new ArrayList<>();
        List<Pair> max2 = new ArrayList<>();
        for (int i = 0; i <= k; i++) {
            max1.add(new Pair(0, null));
            max2.add(new Pair(0, null));
        }

        for (int num : nums) {
            int[] new_lens_for_num = new int[k + 1];
            for (int j = 0; j <= k; j++) {
                int len_same = 1 + dp.get(j).getOrDefault(num, 0);

                int len_diff = 0;
                if (j > 0) {
                    Pair m1 = max1.get(j - 1);
                    Integer m1_val = m1.val;
                    if (m1_val == null || m1_val != num) {
                        len_diff = 1 + m1.len;
                    } else {
                        Pair m2 = max2.get(j - 1);
                        len_diff = 1 + m2.len;
                    }
                }

                if (j == 0) {
                    new_lens_for_num[j] = len_same;
                } else {
                    new_lens_for_num[j] = Math.max(len_same, len_diff);
                }
            }

            for (int j = 0; j <= k; j++) {
                int l = new_lens_for_num[j];

                dp.get(j).put(num, l);

                Pair m1 = max1.get(j);
                Pair m2 = max2.get(j);

                if (l > m1.len) {
                    if (m1.val == null || num != m1.val) {
                        max2.set(j, new Pair(m1.len, m1.val));
                    }
                    max1.set(j, new Pair(l, num));
                } else if (l > m2.len && (m1.val == null || num != m1.val)) {
                    max2.set(j, new Pair(l, num));
                }
            }
        }

        int ans = 0;
        for (Pair p : max1) {
            ans = Math.max(ans, p.len);
        }

        return ans;
    }

    static class Pair {
        int len;
        Integer val;

        public Pair(int len, Integer val) {
            this.len = len;
            this.val = val;
        }
    }
}

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Problem Info

Description

Problem 3176: Find the Maximum Length of a Good Subsequence I - Detailed explanation and solution approaches.

Topics

Array Hash Table Dynamic Programming

Companies

Snowflake

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3176

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

6 steps

1
Step 1: Initialize DP and Max Length Arrays: Create a 2D DP array `dp` where `dp[j][num]` represents the maximum length of a good subsequence ending with `num` and using at most `j` differences. Also initialize two arrays `max1` and `max2` to store the two maximum lengths of the subsequences we have seen so far, and their respective ending element values. Note that each row of 'dp' is a dictionary.
2
Step 2: Iterate Through the Input Array: Iterate through the input array `nums`, element by element.
3
Step 3: Calculate Lengths with Same and Different Numbers: For each `num` in `nums` and for each allowed difference count `j` (from 0 to `k`), calculate two potential lengths: `len_same` (length of subsequence if the current `num` is the same as the last element of the subsequence) and `len_diff` (length if the current `num` is different).
4
Step 4: Update Lengths: Determine the new max length given the current num. Store it in 'new_lens_for_num'.
5
Step 5: Update DP and Max Length Arrays: Update the dp table and the max arrays using the new max length.
6
Step 6: Find Maximum Length: After iterating through the entire input array, find the maximum length subsequence possible by checking the maximum values stored in the max length array.

💡

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(nk)

💾

Space Complexity

O(nk)

📝

Why?

The nested loops in the algorithm lead to the O(nk) time complexity. The `dp` list of dictionaries, the `max1`, and `max2` lists of pairs each contributes a factor of `k`. Thus, they contribute to the overall space complexity of O(nk).

📖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 check if each one is a good subsequence based on the k constraint, then track the maximum length seen.

Time: O(2^n * n), where n is the number of elements in `nums`. Generating subsequences is O(2^n), and checking each subsequence for 'goodness' and finding the length is O(n).Space: O(n), for storing subsequence.

Trade-offs: Brute force is much simpler to understand but highly inefficient for large input sizes. It's useful for understanding the problem but not practical for larger constraints. Dynamic programming is a better approach.

✨

Key Insights

1Insight 1: Dynamic Programming: The problem's optimal substructure makes dynamic programming a suitable approach. We can build up the solution by considering smaller subproblems.
2Insight 2: State Representation: A key challenge is representing the state in the DP. We need to track the subsequence length and the number of allowed differences (k).
3Insight 3: Optimization: The original approach keeps track of the maximum length good subsequence. Because we only need to find the maximum length subsequence, the 'max1' and 'max2' arrays are used to optimize the DP process.

📚

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: If the input array `nums` is empty, the algorithm correctly returns 0, as there are no subsequences to consider. This is handled at the very beginning of the code.
Edge case 2: k is 0: When k is zero, the good subsequence can only have identical adjacent elements. The algorithm correctly handles this scenario by only allowing the 'len_same' logic to run and keeping track of the longest possible subsequence made from identical numbers.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect DP State Definition: Defining the DP state incorrectly can lead to an inefficient or incorrect algorithm. Understanding how to incorporate 'k' into the state is crucial.

✗Common mistake 2: Forgetting to Handle k=0 Case Correctly: Forgetting to properly handle edge cases, such as when k=0, can result in incorrect output.

✅

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 subsequences with minimal differences can be useful in data compression algorithms to identify patterns and reduce redundancy.

🏦Real-world scenario 2: Bioinformatics: Analyzing DNA sequences often involves finding subsequences with certain properties. The 'k' constraint can represent the number of mutations or errors allowed in the sequence alignment.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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 25, 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 13

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

@InterviewProJul 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 are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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

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

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