Longest String Chain

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.
SOLUTION:
from collections import defaultdict

class Solution:
    def longest(self, graph, w, longestCache):
        if w in longestCache:
            return longestCache[w]
        mlen = 1
        for succ in graph[w]:
            l = self.longest(graph, succ, longestCache)
            mlen = max(mlen, 1 + l)
        longestCache[w] = mlen
        return mlen
    
    def longestStrChain(self, words: List[str]) -> int:
        n = len(words)
        graph = defaultdict(list)
        words = set(words)
        for w in words:
            wlen = len(w)
            for c in 'abcdefghijklmnopqrstuvwxyz':
                for i in range(wlen + 1):
                    s = w[:i] + c + w[i:]
                    if s in words:
                        graph[w].append(s)
        longestCache = {}
        mlen = 1
        for w in words:
            l = self.longest(graph, w, longestCache)
            mlen = max(mlen, l)
        return mlen

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