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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.

Solution:

Algorithm:

  • Sort words by word length in ascending order
  • Initialize Hashmap H of type string->int
  • For each word find subwords (word-1 char) and check if subword is in H. Find the maximum possible value of H[subword] out of all subwords. Call it ctr. If there is no subword in H then ctr=0
  • H[word] = ctr+1
  • Finally answer is the maximum of all values in H.
 
bool complen(string s1, string s2)
{
    return s1.length() < s2.length();
}

class Solution {
public:
    int longestStrChain(vector<string>& words) {
        unordered_map<string,int> H;
        int res=0;
        sort(words.begin(), words.end(), complen);
        for (string word: words)
        {
            int wn = word.length(), ctr=0;
            for (int i=0; i<wn; i++)
            {
                string sub = string(word.begin(),word.begin()+i) + 
                             string(word.begin()+i+1, word.end());
                if (H.count(sub)) ctr=max(ctr,H[sub]);
            }
            H[word]=ctr+1;
            res = max(res, H[word]);
        }
        return res;
    }
};
 
 
TC: O(N*|S| + N*log(N))
SC: O(N)

where N=#(words), |S|=max length of word

Reasoning for TC: N*|S| term comes because for each word we are finding all the subwords N*log(N) term comes because we are doing a sorting operation in the beginning

Reasoning for SC: Size of HashMap is N.

PLEASE UPVOTE.