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.

Examples

Example 1:

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Input: words = ["a","b","ba","bca","bda","bdca"]
Output: 4
Explanation: One of the longest word chains is ["a","ba","bda","bdca"].

Example 2:

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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:

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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.

Solution

Method 1 - DP

  • Sort the words by word’s length. (also can apply bucket sort)
  • For each word, generate all possible previous word prev with 1 letter missing.
  • If we have seen this previous word, update the longest chain for the current word.
  • Finally return the longest word chain.

Here is the approach:

  1. Sorting: Sort the words by their lengths. This ensures that when we try to build the word chain, we only look at words with lengths less than the current word.
  2. Dynamic Programming: Use a hashmap to keep track of the longest chain ending at each word.
  3. Checking Predecessors: For each word, try to remove each character one by one to see if the resulting word exists in the hashmap, indicating a valid predecessor.

Code

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class Solution {
    public int longestStrChain(String[] words) {
        Arrays.sort(words, (a, b) -> a.length() - b.length());
        Map<String, Integer> dp = new HashMap<>();
        int ans = 0;

        for (String word : words) {
            int maxLen = 0;
            for (int i = 0; i < word.length(); i++) {
                String predecessor = word.substring(0, i) + word.substring(i + 1);
                maxLen = Math.max(maxLen, dp.getOrDefault(predecessor, 0) + 1);
            }
            dp.put(word, maxLen);
            ans = Math.max(ans, maxLen);
        }

        return ans;
    }
}
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class Solution:
    def longestStrChain(self, words: List[str]) -> int:
        words.sort(key=len)
        dp = {}
        ans = 0
        
        for word in words:
            max_len = 0
            for i in range(len(word)):
                predecessor = word[:i] + word[i+1:]
                max_len = max(max_len, dp.get(predecessor, 0) + 1)
            dp[word] = max_len
            ans = max(ans, max_len)
        
        return ans

Complexity

  • ⏰ Time complexity: O(n * m^2), where n is the number of words, and m is the maximum length of a word. The sorting takes O(n log n). For each word, we potentially look at m predecessor candidates, each taking O(m) to check.
  • 🧺 Space complexity: O(n), used by the hashmap to store the longest chain lengths of words.