Longest String Chain | Leetcode 1048 solution

on

 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 : 


class Solution {
public:
   static bool solve(string &a , string &b)
   {
       return a.length() < b.length();
   }
    bool compare(string &a , string &b )
    {
        if(a.length() < b.length())
        {
            return compare(b , a);
        }
        if(a.length() != b.length() +1)
        {
            return false;
        }
        int flag =0;
        int j =0;
        int size = a.size();
        for(int i =0 ; i < size ;i++)
        {
            if(a[i] != b[j])
            {   
                if(flag ==1)
                {
                    return false;
                }
                 flag =1;
                 continue;
            }
            else
            {
                 j++;
            }

        }
        if( j == b.length())
        {
            return true;
        }
        return false;
    }
    int longestStrChain(vector<string>& words) {
        
          int cnt = INT_MIN;
          sort(words.begin() , words.end(), solve);
          int size = words.size();
          vector<int>dp(size +1 , 1);
          for(int i = 0; i < size ; i++)
          {
              for(int j = 0 ; j < i ; j++)
              {
                  if(compare(words[i], words[j]) )
                  {
                      dp[i] = max(dp[i], dp[j] +1);
                  }
              }
              cnt = max(cnt , dp[i]);
          }
          return cnt;
    }
};

Comments

Post a Comment