Introduction

In this blog post, we will explore how to solve the LeetCode problem "1048" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: Utilize dynamic programming to store the length of the longest word chain ending at each word. The core idea is to build the chain incrementally.
• Length-Based Processing: Process words in order of their length. This ensures that when considering a word, all its potential predecessors have already been processed.

• Predecessor Check: Efficiently check if one word is a predecessor of another by iterating through the characters and allowing for a single insertion.

• Runtime Complexity: O(N * L^2), where N is the number of words and L is the maximum length of a word.

• Storage Complexity: O(N), where N is the number of words.

Code

    def longestStrChain(words):
    word_set = set(words)
    dp = {}
    max_length = 0

    def find_chain_length(word):
        if word in dp:
            return dp[word]

        max_chain = 1  # Base case: A single word is a chain of length 1

        for i in range(len(word)):
            predecessor = word[:i] + word[i+1:]
            if predecessor in word_set:
                max_chain = max(max_chain, find_chain_length(predecessor) + 1)

        dp[word] = max_chain
        return max_chain

    for word in words:
        max_length = max(max_length, find_chain_length(word))

    return max_length