一、题目

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

二、题解

枚举只包含1种字符，满足大于等于k个重复字符的最大字符长度。再枚举包含2种字符，满足大于等于k个重复字符的最大字符长度。一直到包含26种字符，满足大于等于k个重复字符的最大字符长度。

class Solution {
public:int longestSubstring(string s, int k) {int res = 0,n = s.size();unordered_map<char,int> cnt;//枚举1-26的情况for(int req = 1;req <= 26;req++){cnt.clear();for(int l = 0,r = 0,satisfy = 0,collect = 0;r < n;r++){cnt[s[r]]++;if(cnt[s[r]] == 1) collect++;if(cnt[s[r]] == k) satisfy++;while(collect > req){if(cnt[s[l]] == 1) collect--;if(cnt[s[l]] == k) satisfy--;cnt[s[l++]]--;}if(satisfy == req) res = max(res,r - l + 1);}}return res;}
};