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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

Solution 1: Dynamic Programming + Sorting

This implementation uses dynamic programming, followed by sorting. It traverses the relevant values and updates its state as each value is processed. Previously computed states are retained so the same subproblem does not need to be solved again. Ordering the data first makes the subsequent comparisons deterministic.

Solution 2

This implementation uses sorting. It traverses the relevant values and updates its state as each value is processed. Ordering the data first makes the subsequent comparisons deterministic. After all required states have been considered, the maintained result is returned.

  • 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


# Solution 2
class Solution:
    def longestStrChain(self, words: List[str]) -> int:
        words.sort(key=lambda x: len(x))
        res = 0
        mp = {}
        for word in words:
            x = 1
            for i in range(len(word)):
                pre = word[:i] + word[i + 1 :]
                x = max(x, mp.get(pre, 0) + 1)
            mp[word] = x
            res = max(res, x)
        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 {
    #[allow(dead_code)]
    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
    }
}

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