0030. Substring with Concatenation of All Words

https://leetcode.com/problems/substring-with-concatenation-of-all-words

Description

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.

ac1: 2 maps

time O(n k m), k is word count, m is word length space O(2k), k is word count

class Solution {
    public List<Integer> findSubstring(String s, String[] words) {
        List<Integer> res = new ArrayList<Integer>();
        // edge cases
        if (s == null || words.length == 0 || s.length() < words[0].length()) return res;

        // walk words, record word and count in map
        int sl = s.length(), wl = words[0].length(), wcnt = words.length;
        Map<String, Integer> map = new HashMap<String, Integer>();
        for (String str : words) {
            map.put(str, map.containsKey(str) ? map.get(str) + 1 : 1);
        }

        // walk string
        for (int i = 0; i <= sl - wl * wcnt; i++) {
            Map<String, Integer> seen = new HashMap<String, Integer>();
            int j = i;
            int k = wcnt;
            while (k > 0) {
                String w = s.substring(j, j + wl);
                int seencnt = seen.getOrDefault(w, 0);
                if (!map.containsKey(w) || seencnt >= map.get(w)) break;
                seen.put(w, seencnt + 1);
                k--;
                j += wl;
            }

            if (k == 0) res.add(i); // this window has all words
        }

        // result
        return res;
    }
}
/*
1. walk words, record word and count in map

2. walk string:
    copy a map, check whether window meet the requirement
*/