# 1048. Longest String Chain

## Description

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.

## Solutions

• class Solution {
public int longestStrChain(String[] words) {
Arrays.sort(words, Comparator.comparingInt(String::length));
int res = 0;
Map<String, Integer> map = new HashMap<>();
for (String word : words) {
int x = 1;
for (int i = 0; i < word.length(); ++i) {
String pre = word.substring(0, i) + word.substring(i + 1);
x = Math.max(x, map.getOrDefault(pre, 0) + 1);
}
map.put(word, x);
res = Math.max(res, x);
}
return res;
}
}

• class Solution {
public:
int longestStrChain(vector<string>& words) {
sort(words.begin(), words.end(), [&](string a, string b) { return a.size() < b.size(); });
int res = 0;
unordered_map<string, int> map;
for (auto word : words) {
int x = 1;
for (int i = 0; i < word.size(); ++i) {
string pre = word.substr(0, i) + word.substr(i + 1);
x = max(x, map[pre] + 1);
}
map[word] = x;
res = max(res, x);
}
return res;
}
};

• class Solution:
def longestStrChain(self, words: List[str]) -> int:
def check(w1, w2):
if len(w2) - len(w1) != 1:
return False
i = j = cnt = 0
while i < len(w1) and j < len(w2):
if w1[i] != w2[j]:
cnt += 1
else:
i += 1
j += 1
return cnt < 2 and i == len(w1)

n = len(words)
dp = [1] * (n + 1)
words.sort(key=lambda x: len(x))
res = 1
for i in range(1, n):
for j in range(i):
if check(words[j], words[i]):
dp[i] = max(dp[i], dp[j] + 1)
res = max(res, dp[i])
return res


• func longestStrChain(words []string) int {
sort.Slice(words, func(i, j int) bool { return len(words[i]) < len(words[j]) })
res := 0
mp := make(map[string]int)
for _, word := range words {
x := 1
for i := 0; i < len(word); i++ {
pre := word[0:i] + word[i+1:len(word)]
x = max(x, mp[pre]+1)
}
mp[word] = x
res = max(res, x)
}
return res
}

• function longestStrChain(words: string[]): number {
words.sort((a, b) => a.length - b.length);
let ans = 0;
let hashTable = new Map();
for (let word of words) {
let c = 1;
for (let i = 0; i < word.length; i++) {
let pre = word.substring(0, i) + word.substring(i + 1);
c = Math.max(c, (hashTable.get(pre) || 0) + 1);
}
hashTable.set(word, c);
ans = Math.max(ans, c);
}
return ans;
}


• use std::collections::HashMap;

impl Solution {
pub fn longest_str_chain(words: Vec<String>) -> i32 {
let mut words = words;
let mut ret = 0;
let mut map: HashMap<String, i32> = HashMap::new();

// Sort the words vector first
words.sort_by(|lhs, rhs| { lhs.len().cmp(&rhs.len()) });

// Begin the "dp" process
for w in words.iter() {
let n = w.len();
let mut x = 1;

for i in 0..n {
let s = w[..i].to_string() + &w[i + 1..];
let v = map.entry(s.clone()).or_default();
x = std::cmp::max(x, *v + 1);
}

map.insert(w.clone(), x);

ret = std::cmp::max(ret, x);
}

ret
}
}