1.11 Words Concatenation (hard)

Problem:

30. Substring with Concatenation of All Words

You are given a string s and an array of strings words of the same length. Return all starting indices of substring(s) in s that is a concatenation of each word in words exactly once, in any order, and without any intervening characters.

You can return the answer in any order.

Example 1:

Input: s = "barfoothefoobarman", words = ["foo","bar"]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are "barfoo" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.

Example 2:

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

Example 3:

Input: s = "barfoofoobarthefoobarman", words = ["bar","foo","the"]
Output: [6,9,12]

Constraints:

• 1 <= s.length <= 104

• s consists of lower-case English letters.

• 1 <= words.length <= 5000

• 1 <= words[i].length <= 30

• words[i] consists of lower-case English letters.

Solution:

class Solution {
    public List<Integer> findSubstring(String s, String[] words) {
        final Map<String, Integer> counts = new HashMap<>();
        for (final String word : words) {
            counts.put(word, counts.getOrDefault(word, 0) + 1);
        }
        final List<Integer> indexes = new ArrayList<>();
        final int n = s.length(), num = words.length, len = words[0].length();
        for (int i = 0; i < n - num * len + 1; i++) {
            final Map<String, Integer> seen = new HashMap<>();
            int j = 0;
            while (j < num) {
                final String word = s.substring(i + j * len, i + (j + 1) * len);
                if (counts.containsKey(word)) {
                    seen.put(word, seen.getOrDefault(word, 0) + 1);
                    if (seen.get(word) > counts.getOrDefault(word, 0)) {
                        break;
                    }
                } else {
                    break;
                }
                j++;
            }
            if (j == num) {
                indexes.add(i);
            }
        }
        return indexes;
    }
}