README_EN.md

395. Longest Substring with At Least K Repeating Characters

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Description

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.

 

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 <= s.length <= 104
• s consists of only lowercase English letters.
• 1 <= k <= 105

Solutions

Python3

class Solution:
    def longestSubstring(self, s: str, k: int) -> int:
        n = len(s)
        maxLength = 0

        for i in range(1, 27):
            counter = dict()
            left = 0
            right = 0
            unique = 0
            kCount = 0

            while right < n:
                if unique <= i:
                    r = s[right]
                    counter[r] = counter.get(r, 0) + 1

                    if counter[r] == 1:
                        unique += 1
                    if counter[r] == k:
                        kCount += 1

                    right += 1

                else:
                    l = s[left]
                    counter[l] = counter.get(l, 0) - 1

                    if counter[l] == 0:
                        unique -= 1
                    if counter[l] == k - 1:
                        kCount -= 1

                    left += 1

                if unique == i and kCount == i:
                    maxLength = max(maxLength, right - left)

        return maxLength

Java

class Solution {
    public int longestSubstring(String s, int k) {
        int maxLength = 0;
        int n = s.length();

        for (int i = 1; i < 27; i++) {
            Map<Character, Integer> counter = new HashMap<>();
            int left = 0;
            int right = 0;
            int unique = 0;
            int kCount = 0;

            while (right < n) {
                if (unique <= i) {
                    char r = s.charAt(right);
                    counter.put(r, counter.getOrDefault(r, 0) + 1);
                    if (counter.get(r) == 1) {
                        unique += 1;
                    }
                    if (counter.get(r) == k) {
                        kCount += 1;
                    }
                    right += 1;
                } else {
                    char l = s.charAt(left);
                    counter.put(l, counter.getOrDefault(l, 2) - 1);
                    if (counter.get(l) == 0) {
                        unique -= 1;
                    }
                    if (counter.get(l) == k - 1) {
                        kCount -= 1;
                    }
                    left += 1;
                }
                if (unique == i && kCount == i) {
                    maxLength = Math.max(maxLength, right - left);
                }
            }
        }
        return maxLength;
    }
}

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