30. Substring with Concatenation of All Words

You are given a string s and an array of strings words. All the strings of words are of the same length.

A concatenated string is a string that exactly contains all the strings of any permutation of words concatenated.

• For example, if words = ["ab","cd","ef"], then "abcdef", "abefcd", "cdabef", "cdefab", "efabcd", and "efcdab" are all concatenated strings. "acdbef" is not a concatenated string because it is not the concatenation of any permutation of words.

Return an array of the starting indices of all the concatenated substrings in s. You can return the answer in any order.

Example 1:

Input: s = "barfoothefoobarman", words = ["foo","bar"]

Output: [0,9]

Explanation:

The substring starting at 0 is "barfoo". It is the concatenation of ["bar","foo"] which is a permutation of words. The substring starting at 9 is "foobar". It is the concatenation of ["foo","bar"] which is a permutation of words.

Example 2:

Input: s = "wordgoodgoodgoodbestword", words = ["word","good","best","word"]

Output: []

Explanation:

There is no concatenated substring.

Example 3:

Input: s = "barfoofoobarthefoobarman", words = ["bar","foo","the"]

Output: [6,9,12]

Explanation:

The substring starting at 6 is "foobarthe". It is the concatenation of ["foo","bar","the"]. The substring starting at 9 is "barthefoo". It is the concatenation of ["bar","the","foo"]. The substring starting at 12 is "thefoobar". It is the concatenation of ["the","foo","bar"].

Constraints:

• 1 <= s.length <= 104
• 1 <= words.length <= 5000
• 1 <= words[i].length <= 30
• s and words[i] consist of lowercase English letters.

Solution:

class Solution {
    public List<Integer> findSubstring(String s, String[] words) {
        List<Integer> result = new ArrayList<>();

        if (s.length() == 0 || words.length == 0){
            return result;
        }

        int wordLen = words[0].length();
        int wordCount = words.length;

        int windowSize = wordLen * wordCount;

        Map<String, Integer> wordMap = new HashMap<>();

        for (String word : words){
            wordMap.put(word, wordMap.getOrDefault(word, 0) + 1);
        }
                // 枚举第一个窗口的左端点，做 wordLen 次起点不同的滑动窗口
        for (int i = 0; i < wordLen; i++){
            int left = i;
            int right = i;

            int count = 0;

            Map<String, Integer> windowMap = new HashMap<>();

            while(right + wordLen <= s.length()){
                String word = s.substring(right, right + wordLen);

                right = right + wordLen;

                if (wordMap.containsKey(word)){
                    windowMap.put(word, windowMap.getOrDefault(word, 0) + 1);
                    count++;

                    while(windowMap.get(word) > wordMap.get(word)){

                        String leftWord = s.substring(left, left + wordLen);

                        windowMap.put(leftWord, windowMap.get(leftWord) - 1);

                        left = left + wordLen;
                        count--;
                    }

                    if (count == wordCount){
                        result.add(left);
                    }
                }else{
                    windowMap.clear();
                    count = 0;
                    left = right;
                }
            }
        }

        return result;
    }
}

// TC: O(n)
// SC: O(k)

1456. Maximum Number of Vowels in a Substring of Given Length

567. Permutation in String