395. Longest Substring with At Least K Repeating Characters - Detailed Explanation

Problem Statement

Given a string s and an integer k, find the length of the longest substring of s such that the frequency of every character in this substring is at least k. In other words, every character that appears in the substring must appear at least k times.

Example 1

• Input: s = "aaabb", k = 3
• Output: 3
• Explanation: The longest substring that satisfies the condition is "aaa" where the character 'a' appears 3 times.

Example 2

• Input: s = "ababbc", k = 2
• Output: 5
• Explanation: The longest substring is "ababb" where 'a' appears 2 times and 'b' appears 3 times.

Constraints

• s consists only of lowercase English letters.
• 1 ≤ s.length ≤ 10⁴ (or higher in some versions)
• 1 ≤ k ≤ 10⁴

Hints

1. Frequency Check:
   First, count the frequency of every character in the string. If every character meets the frequency threshold k, then the entire string is the answer.

2. Splitting on Invalid Characters:
   If you find a character whose total frequency is less than k, then this character cannot be part of any valid substring. Use it as a delimiter to split the string and solve for each part recursively.

3. Recursive Divide and Conquer:
   By splitting the string around characters that are “invalid” (appear fewer than k times), you narrow down the problem and search for valid substrings in smaller segments.

Approaches Overview

Approach 1: Brute Force (Inefficient)

• Idea:
  Check every possible substring and validate if each character in that substring appears at least k times.
• Drawbacks:
  With a string of length n, there are O(n²) substrings and checking each substring for frequency takes additional time, leading to an overall inefficient solution for larger inputs.

Approach 2: Divide and Conquer (Recursive Splitting)

• Idea:
  • Count the frequency of each character in the current substring.
  • If all characters satisfy the condition (frequency ≥ k), return the length of the substring.
  • Otherwise, identify a character with frequency less than k. This character cannot be part of any valid solution, so split the string by this character.
  • Recursively solve the problem for each resulting substring and return the maximum length.
• Benefits:
  This method often prunes many unnecessary checks and can run much faster in practice compared to brute force.

Approach 3: Sliding Window (Optimized for Fixed Unique Count)

• Idea:
  Use a sliding window to iterate through the string while maintaining counts for characters and dynamically adjusting the window to satisfy the condition of having a fixed number of unique characters.
• Procedure:
  • Try for all possible numbers of unique characters (from 1 up to 26).
  • Within each iteration, use two pointers to form a window and adjust it while keeping track of the count of characters and the count of characters meeting the k threshold.
• Benefits:
  This approach can be very efficient when the alphabet size is limited.

Detailed Step-by-Step Explanation (Divide and Conquer)

1. Count Frequencies:
   For the given substring, count how many times each character appears.

2. Check Validity:
   If every character in the substring has a frequency ≥ k, then the substring is valid. Return its length.

3. Find a Splitting Character:
   Iterate over the substring and identify a character whose frequency is less than k. This character can never be part of a valid substring.

4. Split and Recurse:
   Use the identified character as a delimiter to split the substring into smaller segments.
   Recursively call the function on each segment to find the maximum valid substring length.

5. Combine Results:
   The answer is the maximum length found among all segments.

Python Implementation (Divide and Conquer)