0395. Longest Substring with At Least K Repeating Characters (M)

Longest Substring with At Least K Repeating Characters (M)

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 less 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 <= 10^4
• s consists of only lowercase English letters.
• 1 <= k <= 10^5

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题意

在指定字符串中找到一个最长的子串，使其包含的每一个字母出现的次数都大于等于一个阈值。

思路

求子串问题通常都会想到滑动窗口，但由于本题是要求超过阈值，直接用滑动窗口很难想出一个用来判断该缩短仍是伸长窗口的标准。官方解答提供了一个很巧妙的思路：一个子串能拥有的不一样字符的种类是受原字符串限制的，因此能够将窗口变更的依据定为“当前子串拥有的不一样字符的个数”，这样最多进行26次遍历便可获得答案。

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代码实现

Java

class Solution {
    public int longestSubstring(String s, int k) {
        int ans = 0;
        int uniqueCount = 0;
        boolean[] exist = new boolean[26];

        for (char c : s.toCharArray()) {
            if (!exist[c - 'a']) {
                exist[c - 'a'] = true;
                uniqueCount++;
            }
        }

        for (int ceil = 1; ceil <= uniqueCount; ceil++) {
            int unique = 0, satisfied = 0, start = 0, end = 0;
            int[] count = new int[26];

            while (end < s.length()) {
                if (unique <= ceil) {
                    int index = s.charAt(end) - 'a';
                    if (count[index] == 0) unique++;
                    count[index]++;
                    if (count[index] == k) satisfied++;
                    end++;
                } else {
                    int index = s.charAt(start) - 'a';
                    if (count[index] == k) satisfied--;
                    count[index]--;
                    if (count[index] == 0) unique--;
                    start++;
                }

                if (unique == ceil && unique == satisfied) {
                    ans = Math.max(ans, end - start);
                }
            }
        }
        
        return ans;
    }
}