1048 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.

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class Solution:
    def longestStrChain(self, words: List[str]) -> int:
        count = defaultdict(set)
        dp = {}
        min_len = math.inf
        max_len = -math.inf
        for w in words:
            word_len = len(w)
            count[word_len].add(w)
            min_len = min(min_len, word_len)
            max_len = max(max_len, word_len)
        
        for w in count[min_len]:
            dp[w] = 1
        
        for l in range(min_len+1, max_len+1):
            for w in count[l]:
                chain_len = 1
                for i in range(len(w)):
                    pw = w[:i] + w[i+1:]
                    if pw in dp and dp[pw] + 1 > chain_len:
                        chain_len = dp[pw] + 1
                dp[w] = chain_len
        return max(dp.values())


'''
Shorter
'''
class Solution:
    def longestStrChain(self, words: List[str]) -> int:
        dp = {}
        for w in sorted(words, key=len):
            dp[w] = max(
                dp.get(w[:i] + w[i + 1:], 0) + 1
                for i in range(len(w))
            )
        return max(dp.values())