Longest String Chain

You are given an array of lowercase words words representing a dictionary.
A string chain starts with one dictionary word. At each step, remove exactly one character from the current word. The resulting word must also be present in the dictionary.
Return the maximum number of words in any valid chain. A single dictionary word forms a chain of length 1.
FastPrep practice interpretation: Repeated copies of the same word do not create extra chain positions; membership is determined by distinct word values.

Function
longestChain(words: String[]) → int

Examples
Example 1
words = ["a","and","an","bear"]
return = 3
The chain ["and", "an", "a"] removes one character at each step and has length 3.
Example 2
words = ["a","b","ba","bca","bda","bdca"]
return = 4
One longest chain is ["bdca", "bda", "ba", "a"].
Example 3
words = ["abcd","dbqca"]
return = 1
Neither word becomes the other by deleting one character, so the longest chain contains one word.

Constraints
1 <= words.length <= 50000
1 <= words[i].length <= 60
Every word contains only lowercase English letters from a through z.

The trick: sort words by length, then process shortest to longest. For each word, try deleting each character to build a predecessor. Look that predecessor up in a hash map of word to best chain length. Set dp[word] = max(dp[word], dp[pred] + 1), starting at 1. Track the global max. Cost is roughly n times L squared, where L is at most 60, which is fine for 50000 words. The pitfall is the pairwise comparison approach, which is O(n squared) and dies on large inputs. Another one is forgetting duplicates. A map keyed by word value handles that for free. Also don't assume the chain has to start at a one-letter word. A chain can start at any dictionary word. If you blank on the exact order of processing, StealthCoder can hand you the working solution during the live OA.

If you see this problem in your OA tomorrow, the play is to recognize the pattern in 30 seconds. StealthCoder buys you that recognition.