1048. Longest String Chain

Medium

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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.

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