What is the time complexity of Find Longest Special Substring That Occurs Thrice II?

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

What data structures are used to solve Find Longest Special Substring That Occurs Thrice II?

This problem uses Hash Table, String, Binary Search, Sliding Window, Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Find Longest Special Substring That Occurs Thrice II?

Insight 1: The core idea is to group consecutive identical characters and store their lengths. This helps in efficiently counting the occurrences of special substrings. Insight 2: We need to iterate through each character and consider the possible lengths of special substrings formed by that character.

Which companies ask LeetCode 2982. Find Longest Special Substring That Occurs Thrice II in interviews?

This problem is frequently asked at Rubrik.

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Find Longest Special Substring That Occurs Thrice II - LeetCode 2982 Solution

Find Longest Special Substring That Occurs Thrice II - Complete Solution Guide

Find Longest Special Substring That Occurs Thrice II is LeetCode problem 2982, 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 that consists of lowercase English letters. A string is called special if it is made up of only a single character. For example, the string "abc" is not special, whereas the strings "ddd" , "zz" , and "f" are special. Return the length of the longest special substring of s which occurs at least thrice , or -1 if no special substring occurs at least thrice . A substring is a contiguous non-empty sequence of characters within a string. Example 1: Input: s = "aaaa" Output:

Detailed Explanation

The problem asks us to find the length of the longest "special" substring (a substring consisting of only one character repeated) that appears at least three times within a given string `s`. If no such substring exists, we should return -1. A substring must be contiguous and non-empty. The input string `s` consists of lowercase English letters.

Solution Approach

The solution involves first grouping consecutive identical characters together. For each distinct character, we store a list of the lengths of its consecutive occurrences in the string. Then, for each character, we find the three largest lengths among its occurrences (l1, l2, and l3, where l1 >= l2 >= l3). Finally, we calculate the maximum possible length of a special substring that occurs at least three times based on these three lengths. We iterate through all distinct characters and return the maximum length found, or -1 if no suitable special substring is found.

Step-by-Step Algorithm

Step 1: Initialize a data structure (e.g., a hash map or dictionary) to store character groupings and their respective lengths. Keys represent distinct characters present in s, and values represent lists of lengths of consecutive occurrences for that character.
Step 2: Iterate through the input string `s`. For each position, count the length of the current special substring (sequence of identical characters).
Step 3: Store the length of each special substring in the corresponding character group within the hash map.
Step 4: Iterate through the hash map of character groups.
Step 5: For each character group, find the three largest lengths (l1, l2, l3). Finding the top 3 lengths does not require sorting; direct comparison is sufficient.
Step 6: Calculate the potential maximum lengths achievable using these top three lengths. Specifically consider max(l1 - 2, min(l1 - 1, l2), l3).
Step 7: Update the maximum length found so far (ans) with the largest potential maximum length.
Step 8: After processing all characters, return the overall maximum length (ans). If ans remains 0, return -1.
Step 9: Check to see if the answer is still zero, and if so, return -1.

Key Insights

Insight 1: The core idea is to group consecutive identical characters and store their lengths. This helps in efficiently counting the occurrences of special substrings.
Insight 2: We need to iterate through each character and consider the possible lengths of special substrings formed by that character.
Insight 3: The maximum length of a special substring that occurs at least thrice is limited by the lengths of the three longest sequences of the same character. Sorting these sequences for each character is unnecessary; we can find the three largest lengths directly.
Insight 4: Consider the cases where you can form substrings with length l1-2, l1-1, and l1. These cases arise from using the longest substring and creating substrings within it. Using the other lengths allows you to create the necessary counts.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, String, Binary Search, Sliding Window, Counting.

Companies

Asked at: Rubrik.

Find Longest Special Substring That Occurs Thrice II | LeetCode 2982

Medium

Hash Table String Binary Search Sliding Window Counting

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

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(String s) {
        Map<Character, List<Integer>> groups = new HashMap<>();
        
        int i = 0;
        int n = s.length();
        while (i < n) {
            int j = i;
            while (j < n && s.charAt(j) == s.charAt(i)) {
                j += 1;
            }
            int length = j - i;
            if (!groups.containsKey(s.charAt(i))) {
                groups.put(s.charAt(i), new ArrayList<>());
            }
            groups.get(s.charAt(i)).add(length);
            i = j;
        }
        
        int ans = 0;
        
        for (List<Integer> char_lengths : groups.values()) {
            int l1 = 0, l2 = 0, l3 = 0;
            for (int length : char_lengths) {
                if (length > l1) {
                    l3 = l2;
                    l2 = l1;
                    l1 = length;
                } else if (length > l2) {
                    l3 = l2;
                    l2 = length;
                } else if (length > l3) {
                    l3 = length;
                }
            }
            
            if (l1 > 0) {
                ans = Math.max(ans, l1 - 2);
            }
            
            if (l2 > 0) {
                ans = Math.max(ans, Math.min(l1 - 1, l2));
            }
            
            if (l3 > 0) {
                ans = Math.max(ans, l3);
            }
        }
        
        return ans > 0 ? ans : -1;
    }
}

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Problem Info

Description

Problem 2982: Find Longest Special Substring That Occurs Thrice II - Detailed explanation and solution approaches.

Topics

Hash Table String Binary Search Sliding Window Counting

Companies

Rubrik

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2982

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

9 steps

1
Step 1: Initialize a data structure (e.g., a hash map or dictionary) to store character groupings and their respective lengths. Keys represent distinct characters present in s, and values represent lists of lengths of consecutive occurrences for that character.
2
Step 2: Iterate through the input string `s`. For each position, count the length of the current special substring (sequence of identical characters).
3
Step 3: Store the length of each special substring in the corresponding character group within the hash map.
4
Step 4: Iterate through the hash map of character groups.
5
Step 5: For each character group, find the three largest lengths (l1, l2, l3). Finding the top 3 lengths does not require sorting; direct comparison is sufficient.
6
Step 6: Calculate the potential maximum lengths achievable using these top three lengths. Specifically consider max(l1 - 2, min(l1 - 1, l2), l3).
7
Step 7: Update the maximum length found so far (ans) with the largest potential maximum length.
8
Step 8: After processing all characters, return the overall maximum length (ans). If ans remains 0, return -1.
9
Step 9: Check to see if the answer is still zero, and if so, return -1.

💡

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 time complexity is derived from the single pass through the input string `s` for grouping characters. While finding the top three lengths involves a nested loop, the inner loop has a fixed size (maximum of 3 iterations), rendering it constant time. The space complexity stems from the hash map storing lengths of character groups. In the worst-case scenario, the input string alternates between distinct characters, leading to a hash map storing n character groups. Therefore, the space complexity is O(n).

📖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

Sliding Window

Using a sliding window to check substrings of different lengths. For each length, slide the window across the string and count the occurrences of that substring. This is inefficient.

Time: O(n^2) or worseSpace: O(1)

Trade-offs: The sliding window approach is easier to understand conceptually but less efficient than the grouping approach used in the main solution. This would involve a nested loop O(N^2) and checking if the substrings occur three times.

Binary Search

Binary Search over lengths, checking if a certain length occurs thrice. This would work in O(log n) * O(n) time.

Time: O(n log n)Space: O(1)

Trade-offs: Although possibly better than the O(n^2) solution it is inferior to the O(n) solution.

✨

Key Insights

1Insight 1: The core idea is to group consecutive identical characters and store their lengths. This helps in efficiently counting the occurrences of special substrings.
2Insight 2: We need to iterate through each character and consider the possible lengths of special substrings formed by that character.
3Insight 3: The maximum length of a special substring that occurs at least thrice is limited by the lengths of the three longest sequences of the same character. Sorting these sequences for each character is unnecessary; we can find the three largest lengths directly.
4Insight 4: Consider the cases where you can form substrings with length l1-2, l1-1, and l1. These cases arise from using the longest substring and creating substrings within it. Using the other lengths allows you to create the necessary counts.

📚

Key Terms Explained

For Students

Hash Table

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sliding Window

A moving "window" over data to process subsets efficiently.

String

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

Edge Cases

Edge case 1: Input string contains only one distinct character, e.g., "aaaa". The solution correctly identifies the maximum length as 2 because "aa" occurs at least three times.
Edge case 2: Input string contains no repeating consecutive characters, e.g., "abcdef". The solution correctly returns -1 because no special substring occurs thrice.
Edge case 3: Input string is relatively short, like 'aba'. The code needs to be robust enough to handle those cases when the number of identical substrings is not 3 or more.
Edge case 4: Input string contains very long sequences of single characters, such as "aaaaaaaaa". Here, the logic must accurately determine the maximum possible length considering occurrences of 3 or more substrings.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to consider edge cases where the string contains only one distinct character or no repeating characters.

✗Common mistake 2: Incorrectly calculating the number of occurrences of a special substring, leading to inaccurate length determination.

✗Common mistake 3: Assuming the special substring must start at index 0 or end at the last index. All substrings need to be checked regardless of their position.

✗Common mistake 4: Not handling the cases where only one or two sequence lengths are found for a specific character. The code needs to correctly identify l1, l2, and l3 even if some are 0.

✅

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 algorithms can use this approach to find repeating sequences of characters, which can then be encoded more efficiently.

🏦Real-world scenario 2: Bioinformatics can utilize this algorithm to identify repeated sequences of DNA bases within a genome, which may indicate important genetic elements or mutations.

🎮Real-world scenario 3: Log file analysis: Finding frequently repeated error messages or access patterns in log files to identify potential issues or security threats.

💡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.

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 11, 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 13

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

@CodeNinjaJul 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 "Find Longest Special Substring That Occurs Thrice II"?

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.

🤖 Need help with this problem?