# 1048. Longest String Chain

#### Medium

***

You are given an array of `words` where each word consists of lowercase English letters.

`wordA` is a **predecessor** of `wordB` if and only if we can insert **exactly one** letter anywhere in `wordA` **without changing the order of the other characters** to make it equal to `wordB`.

* For example, `"abc"` is a **predecessor** of `"abac"`, while `"cba"` is not a **predecessor** of `"bcad"`.

A **word chain** is a sequence of words `[word1, word2, ..., wordk]` with `k >= 1`, where `word1` is a **predecessor** of `word2`, `word2` is a **predecessor** of `word3`, and so on. A single word is trivially a **word chain** with `k == 1`.

Return *the **length** of the **longest possible word chain** with words chosen from the given list of* `words`.

 

**Example 1:**

```
Input: words = ["a","b","ba","bca","bda","bdca"]
Output: 4
Explanation: One of the longest word chains is ["a","ba","bda","bdca"].
```

**Example 2:**

```
Input: words = ["xbc","pcxbcf","xb","cxbc","pcxbc"]
Output: 5
Explanation: All the words can be put in a word chain ["xb", "xbc", "cxbc", "pcxbc", "pcxbcf"].
```

**Example 3:**

```
Input: words = ["abcd","dbqca"]
Output: 1
Explanation: The trivial word chain ["abcd"] is one of the longest word chains.
["abcd","dbqca"] is not a valid word chain because the ordering of the letters is changed.
```

 

**Constraints:**

* `1 <= words.length <= 1000`
* `1 <= words[i].length <= 16`
* `words[i]` only consists of lowercase English letters.

```python
class Solution:
    def longestStrChain(self, words: List[str]) -> int:
        words.sort(key = lambda x : len(x))
        self.chains = defaultdict(int)
        self.d = {}
        for word in words:
            self.d[word] = 1
        self.longest = 1
        for word in words[::-1]:
            self.dfs(word)
        return self.longest
            
    
    def dfs(self, word, length = 1):
        if word in self.chains:
            return self.chains[word]
        for index in range(len(word)):
            new_word = word[:index] + word[index+1:]
            if new_word in self.d:
                self.dfs(new_word, length+1)
        self.chains[word] = max(self.chains[word], length)
        self.longest = max(self.longest, length)
        return length
```