# 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

```java
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
*/
```


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