LeetCode 솔루션

[6/20] 820. Short Encoding of Words

mingki 쪽지보내기 메일보내기 자기소개 아이디로 검색 전체게시물 작성
작성일 2022.06.19 18:23

8,279 조회
1 댓글
0 추천
0 비추천
목록 게시글 쓰기

Medium

733275Add to ListShare

A valid encoding of an array of words is any reference string s and array of indices indices such that:

words.length == indices.length
The reference string s ends with the '#' character.
For each index indices[i], the substring of s starting from indices[i] and up to (but not including) the next '#' character is equal to words[i].

Given an array of words, return the length of the shortest reference string s possible of any valid encoding of words.

Example 1:

Input: words = ["time", "me", "bell"]
Output: 10
Explanation: A valid encoding would be s = "time#bell#" and indices = [0, 2, 5].
words[0] = "time", the substring of s starting from indices[0] = 0 to the next '#' is underlined in "time#bell#"
words[1] = "me", the substring of s starting from indices[1] = 2 to the next '#' is underlined in "time#bell#"
words[2] = "bell", the substring of s starting from indices[2] = 5 to the next '#' is underlined in "time#bell#"

Example 2:

Input: words = ["t"]
Output: 2
Explanation: A valid encoding would be s = "t#" and indices = [0].

Constraints:

1 <= words.length <= 2000
1 <= words[i].length <= 7
words[i] consists of only lowercase letters.

#Amazon, #HBO, #Array, #String, #Hash Table, #Trie

링크

https://leetcode.com/problems/short-encoding-of-words/ 879 회 연결

익명
작성일 2022.06.20 20:19

Runtime: 394 ms, faster than 26.61% of Python3 online submissions for Short Encoding of Words.
Memory Usage: 17.4 MB, less than 27.98% of Python3 online submissions for Short Encoding of Words.

class TrieNode:
    def __init__(self):
        self.children = {}
class Solution:
    def minimumLengthEncoding(self, words: List[str]) -> int:
        words = list(set(words))
        words = sorted(words, reverse=True, key=len)
        trie_root = TrieNode()
        ends = []
        
        
        for word in words:
            node = trie_root
            for char in word[::-1]:
                if char not in node.children:
                    node.children[char] = TrieNode()
                node = node.children[char]
            if not node.children: ends.append(len(word)+1)
                
        
        return sum(ends)