720. Longest Word in Dictionary

Given a list of strings words representing an English Dictionary, find the longest word in words that can be built one character at a time by other words in words. If there is more than one possible answer, return the longest word with the smallest lexicographical order.

If there is no answer, return the empty string.

Example 1:

Input: 
words = ["w","wo","wor","worl", "world"]
Output: "world"
Explanation: 
The word "world" can be built one character at a time by "w", "wo", "wor", and "worl".

Example 2:

Input: 
words = ["a", "banana", "app", "appl", "ap", "apply", "apple"]
Output: "apple"
Explanation: 
Both "apply" and "apple" can be built from other words in the dictionary. However, "apple" is lexicographically smaller than "apply".

Note:

  • All the strings in the input will only contain lowercase letters.
  • The length of words will be in the range [1, 1000].
  • The length of words[i] will be in the range [1, 30].
  • Rust Solution

    struct Solution;
    
    use std::collections::BTreeMap;
    
    #[derive(Debug, PartialEq, Eq, Clone, Default)]
    struct Trie {
        children: BTreeMap<char, Trie>,
        end: bool,
    }
    
    impl Trie {
        fn new() -> Self {
            Self::default()
        }
        fn from_words(words: Vec<String>) -> Self {
            let mut trie = Trie::new();
            for word in words {
                trie.insert(word);
            }
            trie
        }
        fn insert(&mut self, s: String) {
            let mut link = self;
            for c in s.chars() {
                link = link.children.entry(c).or_default();
            }
            link.end = true;
        }
        fn dfs(&self, s: &mut String, max: &mut String) {
            if self.end {
                if s.len() > max.len() {
                    *max = s.clone();
                }
            }
            if s.is_empty() || self.end {
                for (&c, child) in self.children.iter() {
                    s.push(c);
                    child.dfs(s, max);
                    s.pop();
                }
            }
        }
    }
    
    impl Solution {
        fn longest_word(words: Vec<String>) -> String {
            let trie = Trie::from_words(words);
            let mut s: String = "".to_string();
            let mut max: String = "".to_string();
            trie.dfs(&mut s, &mut max);
            max
        }
    }
    
    #[test]
    fn test() {
        let words: Vec<String> = vec_string!["a", "banana", "app", "appl", "ap", "apply", "apple"];
        let res = "apple".to_string();
        assert_eq!(Solution::longest_word(words), res);
    }
    

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