What is the time complexity of Longest Ideal Subsequence?

The optimal solution has a time complexity of O(26N).

What data structures are used to solve Longest Ideal Subsequence?

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

What are the key insights for solving Longest Ideal Subsequence?

Insight 1: Dynamic Programming: The problem can be solved using dynamic programming by building up the longest ideal subsequence length for each character in the input string. Insight 2: Limited Search Space: The constraint on adjacent characters (absolute difference <= k) limits the search space when extending the subsequence, allowing for efficient computation.

Which companies ask LeetCode 2370. Longest Ideal Subsequence in interviews?

This problem is frequently asked at MakeMyTrip.

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Longest Ideal Subsequence - LeetCode 2370 Solution

Q: How to solve Longest Ideal Subsequence?

The solution uses dynamic programming to efficiently compute the length of the longest ideal subsequence. It iterates through the input string 's'. For each character, it checks all possible preceding characters that can form an ideal subsequence (i.e., those within a difference of 'k' in alphabetical order). It then finds the maximum length of such a subsequence ending with one of these characters and extends it by including the current character. Finally, it updates the dp array with the length of the longest ideal subsequence ending with the current character. The maximum value in the dp array represents the length of the longest ideal subsequence of the entire string 's'.

Q: Which companies ask LeetCode 2370. Longest Ideal Subsequence in interviews?

This problem is frequently asked at MakeMyTrip.

Longest Ideal Subsequence - Complete Solution Guide

Longest Ideal Subsequence is LeetCode problem 2370, 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 s consisting of lowercase letters and an integer k . We call a string t ideal if the following conditions are satisfied: t is a subsequence of the string s . The absolute difference in the alphabet order of every two adjacent letters in t is less than or equal to k . Return the length of the longest ideal string . A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters. Note

Detailed Explanation

The problem asks us to find the length of the longest ideal subsequence of a given string 's'. A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters. An 'ideal' string is one where the absolute difference between the alphabetical order of any two adjacent characters in the subsequence is less than or equal to a given integer 'k'. The input consists of a string 's' containing lowercase English letters and an integer 'k'. The output is the length of the longest ideal subsequence.

Solution Approach

The solution uses dynamic programming to efficiently compute the length of the longest ideal subsequence. It iterates through the input string 's'. For each character, it checks all possible preceding characters that can form an ideal subsequence (i.e., those within a difference of 'k' in alphabetical order). It then finds the maximum length of such a subsequence ending with one of these characters and extends it by including the current character. Finally, it updates the dp array with the length of the longest ideal subsequence ending with the current character. The maximum value in the dp array represents the length of the longest ideal subsequence of the entire string 's'.

Step-by-Step Algorithm

Step 1: Initialize a DP array 'dp' of size 26 (one for each lowercase letter) with all elements set to 0. This array will store the length of the longest ideal subsequence ending with each letter.
Step 2: Iterate through the input string 's' character by character.
Step 3: For each character 'char', calculate its alphabetical index 'current_char_code' by subtracting 'a' from its ASCII value.
Step 4: Determine the range of possible previous characters that can form an ideal subsequence with the current character. This range is defined by 'lower_bound' = max(0, current_char_code - k) and 'upper_bound' = min(25, current_char_code + k).
Step 5: Iterate through the possible previous characters within the range [lower_bound, upper_bound]. For each 'prev_char_code', find the maximum length of the longest ideal subsequence ending with that character (dp[prev_char_code]).
Step 6: Set the value of dp[current_char_code] to max_len + 1, where max_len is the maximum length found in the previous step. This represents the length of the longest ideal subsequence ending with the current character.
Step 7: After processing all characters in 's', find the maximum value in the 'dp' array. This maximum value is the length of the longest ideal subsequence of 's'.
Step 8: Return the maximum length.

Key Insights

Insight 1: Dynamic Programming: The problem can be solved using dynamic programming by building up the longest ideal subsequence length for each character in the input string.
Insight 2: Limited Search Space: The constraint on adjacent characters (absolute difference <= k) limits the search space when extending the subsequence, allowing for efficient computation.
Insight 3: Alphabet Array as DP Table: Since the characters are lowercase English letters, we can use an array of size 26 to store the longest ideal subsequence ending with each letter.

Complexity Analysis

Time Complexity: O(26N)

Space Complexity: O(1)

Topics

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

Companies

Asked at: MakeMyTrip.

Longest Ideal Subsequence | LeetCode 2370

Medium

Hash Table String Dynamic Programming

Time: O(26N)Space: O(1)

Solution Code

Available in:

class Solution {
    public int longestIdealString(String s, int k) {
        int[] dp = new int[26];
        
        for (char charr : s.toCharArray()) {
            int currentCharCode = charr - 'a';
            
            int maxLen = 0;
            
            int lowerBound = Math.max(0, currentCharCode - k);
            int upperBound = Math.min(25, currentCharCode + k);
            
            for (int prevCharCode = lowerBound; prevCharCode <= upperBound; prevCharCode++) {
                maxLen = Math.max(maxLen, dp[prevCharCode]);
            }
            
            dp[currentCharCode] = maxLen + 1;
        }
        
        int max = 0;
        for(int i = 0; i < 26; i++){
            max = Math.max(max, dp[i]);
        }
        return max;
    }
}

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Problem Info

Description

Problem 2370: Longest Ideal Subsequence - Detailed explanation and solution approaches.

Topics

Hash Table String Dynamic Programming

Companies

MakeMyTrip

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2370

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

8 steps

1
Step 1: Initialize a DP array 'dp' of size 26 (one for each lowercase letter) with all elements set to 0. This array will store the length of the longest ideal subsequence ending with each letter.
2
Step 2: Iterate through the input string 's' character by character.
3
Step 3: For each character 'char', calculate its alphabetical index 'current_char_code' by subtracting 'a' from its ASCII value.
4
Step 4: Determine the range of possible previous characters that can form an ideal subsequence with the current character. This range is defined by 'lower_bound' = max(0, current_char_code - k) and 'upper_bound' = min(25, current_char_code + k).
5
Step 5: Iterate through the possible previous characters within the range [lower_bound, upper_bound]. For each 'prev_char_code', find the maximum length of the longest ideal subsequence ending with that character (dp[prev_char_code]).
6
Step 6: Set the value of dp[current_char_code] to max_len + 1, where max_len is the maximum length found in the previous step. This represents the length of the longest ideal subsequence ending with the current character.
7
Step 7: After processing all characters in 's', find the maximum value in the 'dp' array. This maximum value is the length of the longest ideal subsequence of 's'.
8
Step 8: Return the maximum length.

💡

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(26N)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is derived from the nested loops: iterating through the string and then iterating through a maximum of 26 possible characters to find the maximum length. The space complexity is constant because the DP table size is independent of the input string length and solely dependent on the number of lowercase English alphabets which is 26.

📖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 (Recursion)

Generate all possible subsequences and check if each is an ideal string. Then, return the maximum length among ideal strings.

Time: O(2^N), where N is the length of the string 's'.Space: O(N) due to recursion depth in worst case.

Trade-offs: This is highly inefficient and will result in a Time Limit Exceeded (TLE) error for larger input strings. It's useful conceptually but not practical for solving this problem.

✨

Key Insights

1Insight 1: Dynamic Programming: The problem can be solved using dynamic programming by building up the longest ideal subsequence length for each character in the input string.
2Insight 2: Limited Search Space: The constraint on adjacent characters (absolute difference <= k) limits the search space when extending the subsequence, allowing for efficient computation.
3Insight 3: Alphabet Array as DP Table: Since the characters are lowercase English letters, we can use an array of size 26 to store the longest ideal subsequence ending with each letter.

📚

Key Terms Explained

For Students

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.

String

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

Edge Cases

Edge case 1: Empty string: If the input string 's' is empty, the solution correctly returns 0 because the 'dp' array will remain all zeros, and the maximum of 'dp' will be 0.
Edge case 2: k = 0: If k is 0, only identical characters can be adjacent in the ideal subsequence. The algorithm still correctly identifies this and finds the longest subsequence with only one distinct character.
Edge case 3: k = 25: If k is 25, any adjacent lowercase letters are considered ideal. Therefore, the algorithm will identify the string 's' as the longest ideal string, and return the length of s.
Edge case 4: String with only one character: If the string has only one character, the longest ideal subsequence is the string itself, so the length is 1. The algorithm correctly handles this case.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not understanding the subsequence definition. A subsequence is not necessarily contiguous.

✗Common mistake 2: Failing to consider all possible previous characters within the 'k' constraint. Need to iterate through all valid alphabet indices.

✗Common mistake 3: Incorrectly calculating the alphabetical index of a character. Remember to subtract 'a' from the character's ASCII value.

✗Common mistake 4: Using inefficient search within the k constraint. Solution avoids this by iterating through a maximum of 2k+1 possibilities.

✅

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: DNA Sequencing: Identifying similar sequences in DNA strands, where 'k' represents the allowed difference between bases.

🏦Real-world scenario 2: Text Compression: Finding patterns in text where slight variations are allowed to consider them part of the same 'ideal' pattern, which can then be compressed more effectively.

🎮Real-world scenario 3: Recommendation systems: Building recommendation systems where items are considered similar if their characteristics (represented as character sequences) are within a certain threshold 'k'.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

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

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 1, 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 13

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

@AlgoMasterJul 6

❓

Frequently Asked Questions

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.

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.

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 "Longest Ideal Subsequence"?

For this medium-level Hash Table 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.

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