Problem

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.

Java Solution

class Solution {
    public int longestStrChain(String[] words) {
        HashMap<String, Integer> longestPre = new HashMap<>(); // Including the word itself
        TreeMap<Integer, List<String>> wordsByCharLength = new TreeMap<>();
        int res = 1;
        for (String word : words) {
            wordsByCharLength.putIfAbsent(word.length(), new LinkedList<>());
            wordsByCharLength.get(word.length()).add(word);
        }
        int longest = wordsByCharLength.lastKey();
        for (int i = 1; i <= longest; i++) {
            if (!wordsByCharLength.containsKey(i)) {
                continue;
            }
            for (String curWord : wordsByCharLength.get(i)) {
                if (!wordsByCharLength.containsKey(i - 1)) {
                    longestPre.put(curWord, 1);
                    continue;
                }
                int prevLongest = 0;
                for (String prevWord : wordsByCharLength.get(i - 1)) {
                    if (isPredecessorOf(prevWord, curWord)) {
                        prevLongest = Math.max(prevLongest, longestPre.get(prevWord));
                    }
                }
                longestPre.put(curWord, prevLongest + 1);
                res = Math.max(res, prevLongest + 1);
            }
        }
        return res;
    }

    private boolean isPredecessorOf(String prevWord, String curWord) {
        int skipped = 0;
        int i = 0;
        while (i < prevWord.length() && i + skipped < curWord.length()) {
            if (prevWord.charAt(i) != curWord.charAt(i + skipped)) {
                if (skipped == 0) {
                    skipped = 1;
                    continue;
                } else {
                    return false;
                }
            }
            i++;
        }
        return true;
    }
}