Medium
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
.
"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.class Solution {
fun longestStrChain(words: Array<String>): Int {
val lenStr = arrayOfNulls<MutableList<String>?>(20)
for (word in words) {
val len = word.length
if (lenStr[len] == null) {
lenStr[len] = ArrayList()
}
lenStr[len]!!.add(word)
}
val longest: MutableMap<String?, Int?> = HashMap()
var max = 0
for (s in words) {
max = findLongest(s, lenStr, longest).coerceAtLeast(max)
}
return max
}
private fun findLongest(
word: String,
lenStr: Array<MutableList<String>?>,
longest: MutableMap<String?, Int?>,
): Int {
if (longest.containsKey(word)) {
return longest[word]!!
}
val len = word.length
val words: List<String>? = lenStr[len + 1]
if (words == null) {
longest[word] = 1
return 1
}
var max = 0
var i: Int
var j: Int
for (w in words) {
i = 0
j = 0
while (i < len && j - i <= 1) {
if (word[i] == w[j++]) {
++i
}
}
if (j - i <= 1) {
max = findLongest(w, lenStr, longest).coerceAtLeast(max)
}
}
++max
longest[word] = max
return max
}
}