Longest String Chain
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:
We will be using the DFS approach to see all possible substrings in the wordlist to get the length of the longest string chain. We are using memoization to avoid the duplicate calls to for already visited subchains.
- Time Complexity:
O(m*n), n is len(words) and m is len(longest word)
- Space Complexity:
O(n)
for both the wordSet and memo, soO(n)
Problem Credit: LeetCode