Question

720. Longest Word in Dictionary

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Description

Given an array of strings words representing an English Dictionary, return _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.

Note that the word should be built from left to right with each additional character being added to the end of a previous word. 

 

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".

 

Constraints:

• 1 <= words.length <= 1000
• 1 <= words[i].length <= 30
• words[i] consists of lowercase English letters.

Solutions

Solution 1

class Solution:
  def longestWord(self, words: List[str]) -> str:
    cnt, ans = 0, ''
    s = set(words)
    for w in s:
      n = len(w)
      if all(w[:i] in s for i in range(1, n)):
        if cnt < n:
          cnt, ans = n, w
        elif cnt == n and w < ans:
          ans = w
    return ans

class Solution {
  private Set<String> s;

  public String longestWord(String[] words) {
    s = new HashSet<>(Arrays.asList(words));
    int cnt = 0;
    String ans = "";
    for (String w : s) {
      int n = w.length();
      if (check(w)) {
        if (cnt < n) {
          cnt = n;
          ans = w;
        } else if (cnt == n && w.compareTo(ans) < 0) {
          ans = w;
        }
      }
    }
    return ans;
  }

  private boolean check(String word) {
    for (int i = 1, n = word.length(); i < n; ++i) {
      if (!s.contains(word.substring(0, i))) {
        return false;
      }
    }
    return true;
  }
}

class Solution {
public:
  string longestWord(vector<string>& words) {
    unordered_set<string> s(words.begin(), words.end());
    int cnt = 0;
    string ans = "";
    for (auto w : s) {
      int n = w.size();
      if (check(w, s)) {
        if (cnt < n) {
          cnt = n;
          ans = w;
        } else if (cnt == n && w < ans)
          ans = w;
      }
    }
    return ans;
  }

  bool check(string& word, unordered_set<string>& s) {
    for (int i = 1, n = word.size(); i < n; ++i)
      if (!s.count(word.substr(0, i)))
        return false;
    return true;
  }
};

func longestWord(words []string) string {
  s := make(map[string]bool)
  for _, w := range words {
    s[w] = true
  }
  cnt := 0
  ans := ""
  check := func(word string) bool {
    for i, n := 1, len(word); i < n; i++ {
      if !s[word[:i]] {
        return false
      }
    }
    return true
  }
  for w, _ := range s {
    n := len(w)
    if check(w) {
      if cnt < n {
        cnt, ans = n, w
      } else if cnt == n && w < ans {
        ans = w
      }
    }
  }
  return ans
}

function longestWord(words: string[]): string {
  words.sort((a, b) => {
    const n = a.length;
    const m = b.length;
    if (n === m) {
      return a < b ? -1 : 1;
    }
    return m - n;
  });
  for (const word of words) {
    let isPass = true;
    for (let i = 1; i <= word.length; i++) {
      if (!words.includes(word.slice(0, i))) {
        isPass = false;
        break;
      }
    }
    if (isPass) {
      return word;
    }
  }
  return '';
}

impl Solution {
  pub fn longest_word(mut words: Vec<String>) -> String {
    words.sort_unstable_by(|a, b| (b.len(), a).cmp(&(a.len(), b)));
    for word in words.iter() {
      let mut is_pass = true;
      for i in 1..=word.len() {
        if !words.contains(&word[..i].to_string()) {
          is_pass = false;
          break;
        }
      }
      if is_pass {
        return word.clone();
      }
    }
    String::new()
  }
}