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395. 至少有 K 个重复字符的最长子串

English Version

题目描述

给你一个字符串 s 和一个整数 k ，请你找出 s 中的最长子串， 要求该子串中的每一字符出现次数都不少于 k 。返回这一子串的长度。

 

示例 1：

输入：s = "aaabb", k = 3
输出：3
解释：最长子串为 "aaa" ，其中 'a' 重复了 3 次。

示例 2：

输入：s = "ababbc", k = 2
输出：5
解释：最长子串为 "ababb" ，其中 'a' 重复了 2 次， 'b' 重复了 3 次。

 

提示：

• 1 <= s.length <= 104
• s 仅由小写英文字母组成
• 1 <= k <= 105

解法

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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