395. Longest Substring with At Least K Repeating Characters

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Recursion

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class Solution {
    public int longestSubstring(String s, int k) {
        if (s.length() < k) {
            return 0;
        }

        int[] countMap = new int[26];
        for (int i = 0; i < s.length(); i++) {
            countMap[s.charAt(i) - 'a']++;
        }

        for (int i = 0; i < countMap.length; i++) {
            if (countMap[i] > 0 && countMap[i] < k) {
                int maxLength = 0;
                for (String subStr : s.split(String.valueOf((char) (i + 'a')))) {
                    maxLength = Math.max(maxLength, longestSubstring(subStr, k));
                }
                // 注意此处 return 的细节，当前字符不足 k 个时，当前字符不属于任何一个最长字符串，故最长字符串一定在 split 后的子字符串中，此处 return 不会错过最长字符串且减少了运算量
                return maxLength;
            }
        }

        return s.length();
    }
}

Sliding Window

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class Solution {
    public int longestSubstring(String s, int k) {
        if (s.length() < k) {
            return 0;
        }

        int maxLength = 0;
        for (int targetDistinctCharCount = 1; targetDistinctCharCount <= 26; targetDistinctCharCount++) {
            int[] windowMap = new int[26];
            int distinctCharCount = 0;
            int kCharCount = 0;
            int i = 0;
            for (int j = 0; j < s.length(); j++) {
                // window: [i, j]
                char rightmost = s.charAt(j);
                if (windowMap[rightmost - 'a'] == 0) {
                    distinctCharCount++;
                }
                if (++windowMap[rightmost - 'a'] == k) {
                    kCharCount++;
                }

                while (distinctCharCount > targetDistinctCharCount) {
                    char leftmost = s.charAt(i++);
                    if (windowMap[leftmost - 'a']-- == k) {
                        kCharCount--;
                    }
                    if (windowMap[leftmost - 'a'] == 0) {
                        distinctCharCount--;
                    }
                }

                if (kCharCount == targetDistinctCharCount) {
                    maxLength = Math.max(maxLength, j - i + 1);
                }
            }
        }

        return maxLength;
    }
}

此题与常规滑动窗口思路不同的点在于若不固定用到的不同字符的个数，则无法判断何时收缩窗口，因为收缩窗口与求最长的长度是相悖的。当固定了用到的不同字符的个数时，就能明确收缩窗口的目标。

References

395. Longest Substring with At Least K Repeating Characters