#+ANKI_DECK: study_deck_02
* TODO 1408. String Matching in an Array :easy:
:PROPERTIES:
:NEETCODE: [[file:../../roadmap.org::*1408. String Matching in an Array][1408. String Matching in an Array]]
:END:

Given an array of string ~words~, return all strings in/ /~words~/ /that are a substring of another word. You can return the answer in *any order*.

*Example 1:*


#+begin_src
Input: words = ["mass","as","hero","superhero"]
Output: ["as","hero"]
Explanation: "as" is substring of "mass" and "hero" is substring of "superhero".
["hero","as"] is also a valid answer.
#+end_src


*Example 2:*


#+begin_src
Input: words = ["leetcode","et","code"]
Output: ["et","code"]
Explanation: "et", "code" are substring of "leetcode".
#+end_src


*Example 3:*


#+begin_src
Input: words = ["blue","green","bu"]
Output: []
Explanation: No string of words is substring of another string.
#+end_src


*Constraints:*

 - ~1 <= words.length <= 100~

 - ~1 <= words[i].length <= 30~

 - ~words[i]~ contains only lowercase English letters.

 - All the strings of ~words~ are *unique*.

** TODO Approach
Write your approach here.

** TODO Python
#+begin_src python
from collections import defaultdict as dd
class Solution:
    def stringMatching(self, words: List[str]) -> List[str]:
        words = reversed(sorted(words))
        def squash(word, trie):
            lst = []
            for c in word:
                lst.append([trie, ''])
                for i in range(len(lst)):
                    lst[i][0] = lst[i][0][c]
                    lst[i][1] += c
                    if not lst[i][0]['ads']:
                        lst[i][0]['ads'] = {lst[i][1]: [word]}
                    else:
                        lst[i][0]['ads'][lst[i][1]].append(word)
        def in_trie(word, trie):
            for c in word:
                if c not in trie:
                    return []
                trie = trie[c]
            return [(word, bw) for bw in trie['ads'][word]]
        trie_maker = lambda: dd(trie_maker)
        trie = trie_maker()
        ans = []
        for s in words:
            ans += in_trie(s, trie)
            squash(s, trie)
        return ans


#+end_src

**

** TODO C++
#+begin_src cpp
class Solution {
public:
    vector<string> stringMatching(vector<string>& words) {
        
    }
};
#+end_src