What is the time complexity of Longest Substring with At Least K Repeating Characters?

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

What data structures are used to solve Longest Substring with At Least K Repeating Characters?

This problem uses Hash Table, String, Divide and Conquer, Sliding Window. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Longest Substring with At Least K Repeating Characters?

Insight 1: The problem can be approached using a sliding window. However, a simple sliding window isn't directly applicable because we need to ensure *all* characters meet the frequency requirement. Insight 2: The solution iterates through possible numbers of unique characters (from 1 to the total number of unique characters in the string). For each possible number of unique characters, it uses a sliding window to find the longest substring containing that number of unique characters, where each of those characters appears at least k times.

Which companies ask LeetCode 395. Longest Substring with At Least K Repeating Characters in interviews?

This problem is frequently asked at Baidu, Yandex.

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Longest Substring with At Least K Repeating Characters - LeetCode 395 Solution

Longest Substring with At Least K Repeating Characters - Complete Solution Guide

Longest Substring with At Least K Repeating Characters is LeetCode problem 395, 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

Given a string s and an integer k , return the length of the longest substring of s such that the frequency of each character in this substring is greater than or equal to k . if no such substring exists, return 0. Example 1: Input: s = "aaabb", k = 3 Output: 3 Explanation: The longest substring is "aaa", as 'a' is repeated 3 times. Example 2: Input: s = "ababbc", k = 2 Output: 5 Explanation: The longest substring is "ababb", as 'a' is repeated 2 times and 'b' is repeated 3 times. Constraints: 1

Detailed Explanation

The problem asks us to find the longest substring within a given string 's' such that every character in that substring appears at least 'k' times. If no such substring exists, we should return 0. The input consists of a string 's' and an integer 'k'. The string contains only lowercase English letters, and the integer 'k' represents the minimum frequency a character must have in a valid substring. The output is the length of the longest valid substring.

Solution Approach

The solution employs a sliding window technique combined with an outer loop that iterates through possible numbers of unique characters in the substring. For each possible number of unique characters, the sliding window expands and contracts to maintain a substring with that specific number of unique characters. During the window's movement, it checks if all the characters within the window appear at least 'k' times. If a substring meets both conditions (the desired number of unique characters and each character appearing at least 'k' times), the solution updates the maximum length found so far.

Step-by-Step Algorithm

Step 1: Calculate the number of unique characters in the input string 's'. This value will limit the maximum number of unique characters we need to consider in our substrings.
Step 2: Iterate through possible numbers of unique characters ('h') in the substring, from 1 to the total number of unique characters.
Step 3: Initialize a sliding window with 'left' and 'right' pointers, a character count map, a 'unique_count' variable (number of unique characters in the window), and an 'at_least_k_count' variable (number of unique characters appearing at least 'k' times in the window).
Step 4: Expand the sliding window by moving the 'right' pointer one step at a time. Update the character count map, 'unique_count', and 'at_least_k_count' accordingly.
Step 5: While the number of unique characters in the window ('unique_count') exceeds the current target number of unique characters ('h'), contract the sliding window by moving the 'left' pointer one step at a time. Update the character count map, 'unique_count', and 'at_least_k_count' accordingly.
Step 6: If the number of unique characters in the window ('unique_count') equals 'h' and the number of characters appearing at least 'k' times ('at_least_k_count') also equals 'h', then update the maximum length found so far.
Step 7: After iterating through all possible 'right' pointer positions, repeat steps 3-6 for the next value of 'h'.
Step 8: Return the maximum length found.

Key Insights

Insight 1: The problem can be approached using a sliding window. However, a simple sliding window isn't directly applicable because we need to ensure *all* characters meet the frequency requirement.
Insight 2: The solution iterates through possible numbers of unique characters (from 1 to the total number of unique characters in the string). For each possible number of unique characters, it uses a sliding window to find the longest substring containing that number of unique characters, where each of those characters appears at least k times.
Insight 3: By considering different numbers of unique characters ('h') and checking if a substring with 'h' unique characters and each character appearing at least 'k' times exists, we can ensure we find the maximum length among all such valid substrings.

Complexity Analysis

Time Complexity: O(N)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, String, Divide and Conquer, Sliding Window.

Companies

Asked at: Baidu, Yandex.

Longest Substring with At Least K Repeating Characters | LeetCode 395

Medium

Hash Table String Divide and Conquer Sliding Window

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;

class Solution {
    public int longestSubstring(String s, int k) {
        int max_len = 0;
        
        Set<Character> uniqueChars = new HashSet<>();
        for (char c : s.toCharArray()) {
            uniqueChars.add(c);
        }
        int num_unique_chars_in_s = uniqueChars.size();
        
        for (int h = 1; h <= num_unique_chars_in_s; h++) {
            Map<Character, Integer> counts = new HashMap<>();
            int left = 0;
            int unique_count = 0;
            int at_least_k_count = 0;
            
            for (int right = 0; right < s.length(); right++) {
                char char_right = s.charAt(right);
                
                if (!counts.containsKey(char_right)) {
                    counts.put(char_right, 0);
                }
                if (counts.get(char_right) == 0) {
                    unique_count += 1;
                }
                counts.put(char_right, counts.get(char_right) + 1);
                if (counts.get(char_right) == k) {
                    at_least_k_count += 1;
                }
                
                while (unique_count > h) {
                    char char_left = s.charAt(left);
                    if (counts.get(char_left) == k) {
                        at_least_k_count -= 1;
                    }
                    counts.put(char_left, counts.get(char_left) - 1);
                    if (counts.get(char_left) == 0) {
                        unique_count -= 1;
                    }
                    left += 1;
                }
                
                if (unique_count == h && at_least_k_count == h) {
                    max_len = Math.max(max_len, right - left + 1);
                }
            }
        }
        
        return max_len;
    }
}

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Problem Info

Description

Problem 395: Longest Substring with At Least K Repeating Characters - Detailed explanation and solution approaches.

Topics

Hash Table String Divide and Conquer Sliding Window

Companies

BaiduYandex

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:395

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: Calculate the number of unique characters in the input string 's'. This value will limit the maximum number of unique characters we need to consider in our substrings.
2
Step 2: Iterate through possible numbers of unique characters ('h') in the substring, from 1 to the total number of unique characters.
3
Step 3: Initialize a sliding window with 'left' and 'right' pointers, a character count map, a 'unique_count' variable (number of unique characters in the window), and an 'at_least_k_count' variable (number of unique characters appearing at least 'k' times in the window).
4
Step 4: Expand the sliding window by moving the 'right' pointer one step at a time. Update the character count map, 'unique_count', and 'at_least_k_count' accordingly.
5
Step 5: While the number of unique characters in the window ('unique_count') exceeds the current target number of unique characters ('h'), contract the sliding window by moving the 'left' pointer one step at a time. Update the character count map, 'unique_count', and 'at_least_k_count' accordingly.
6
Step 6: If the number of unique characters in the window ('unique_count') equals 'h' and the number of characters appearing at least 'k' times ('at_least_k_count') also equals 'h', then update the maximum length found so far.
7
Step 7: After iterating through all possible 'right' pointer positions, repeat steps 3-6 for the next value of 'h'.
8
Step 8: Return the maximum length found.

💡

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)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is derived from the fact that we iterate through the string 's' at most a constant number of times (26 times in the worst case). The space complexity is constant because the character count map stores counts for lowercase English letters only, and the number of such characters is fixed (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

Divide and Conquer

A divide-and-conquer approach recursively splits the string into substrings based on characters that appear less than 'k' times. The longest valid substring is then the maximum of the lengths of the valid substrings in each part.

Time: O(N log N) in the average case, O(N^2) in the worst case.Space: O(N) due to the recursion stack.

Trade-offs: The divide and conquer approach might be more intuitive for some, but the provided solution has a better time complexity of O(N). Divide and Conquer can perform poorly in the worst case, becoming O(N^2).

✨

Key Insights

1Insight 1: The problem can be approached using a sliding window. However, a simple sliding window isn't directly applicable because we need to ensure *all* characters meet the frequency requirement.
2Insight 2: The solution iterates through possible numbers of unique characters (from 1 to the total number of unique characters in the string). For each possible number of unique characters, it uses a sliding window to find the longest substring containing that number of unique characters, where each of those characters appears at least k times.
3Insight 3: By considering different numbers of unique characters ('h') and checking if a substring with 'h' unique characters and each character appearing at least 'k' times exists, we can ensure we find the maximum length among all such valid substrings.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like 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: Empty string. The algorithm correctly returns 0, because the outer loop won't execute.
Edge case 2: String with only one character. If k > 1, it returns 0. If k = 1, it returns 1.
Edge case 3: k is greater than the length of the string. The algorithm will correctly return 0 because the substring can't have characters repeating 'k' times
Edge case 4: k is equal to 1. This means any character is valid. The solution will correctly identify the entire string as a potential substring and return its length.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the number of unique characters exceeds the target number 'h'. This can lead to incorrect substring lengths being calculated.

✗Common mistake 2: Forgetting to update 'at_least_k_count' correctly when adding or removing characters from the window, especially when a character's count reaches or drops below 'k'.

✗Common mistake 3: Failing to initialize the counts map or failing to properly manage the sliding window boundary conditions.

✅

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 validation. This algorithm can be used to validate data to ensure that certain elements or patterns appear a minimum number of times within a dataset.

🏦Real-world scenario 2: Bioinformatics. In analyzing DNA sequences, this algorithm can identify regions where certain nucleotide combinations occur frequently, indicating potentially significant genomic features.

💡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

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

⚠️ 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 6, 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

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

@InterviewProJun 29

❓

Frequently Asked Questions

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.

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 Substring with At Least K Repeating Characters"?

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?