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820. Short Encoding of Words

Description

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.

Solutions

  • class Trie { Trie[] children = new Trie[26]; } class Solution { public int minimumLengthEncoding(String[] words) { Trie root = new Trie(); for (String w : words) { Trie cur = root; for (int i = w.length() - 1; i >= 0; i--) { int idx = w.charAt(i) - 'a'; if (cur.children[idx] == null) { cur.children[idx] = new Trie(); } cur = cur.children[idx]; } } return dfs(root, 1); } private int dfs(Trie cur, int l) { boolean isLeaf = true; int ans = 0; for (int i = 0; i < 26; i++) { if (cur.children[i] != null) { isLeaf = false; ans += dfs(cur.children[i], l + 1); } } if (isLeaf) { ans += l; } return ans; } } 
  • struct Trie { Trie* children[26] = {nullptr}; }; class Solution { public: int minimumLengthEncoding(vector<string>& words) { auto root = new Trie(); for (auto& w : words) { auto cur = root; for (int i = w.size() - 1; i >= 0; --i) { if (cur->children[w[i] - 'a'] == nullptr) { cur->children[w[i] - 'a'] = new Trie(); } cur = cur->children[w[i] - 'a']; } } return dfs(root, 1); } private: int dfs(Trie* cur, int l) { bool isLeaf = true; int ans = 0; for (int i = 0; i < 26; ++i) { if (cur->children[i] != nullptr) { isLeaf = false; ans += dfs(cur->children[i], l + 1); } } if (isLeaf) { ans += l; } return ans; } }; 
  • class Trie: def __init__(self) -> None: self.children = [None] * 26 class Solution: def minimumLengthEncoding(self, words: List[str]) -> int: root = Trie() for w in words: cur = root for i in range(len(w) - 1, -1, -1): idx = ord(w[i]) - ord('a') if cur.children[idx] == None: cur.children[idx] = Trie() cur = cur.children[idx] return self.dfs(root, 1) def dfs(self, cur: Trie, l: int) -> int: isLeaf, ans = True, 0 for i in range(26): if cur.children[i] != None: isLeaf = False ans += self.dfs(cur.children[i], l + 1) if isLeaf: ans += l return ans 
  • type trie struct { children [26]*trie } func minimumLengthEncoding(words []string) int { root := new(trie) for _, w := range words { cur := root for i := len(w) - 1; i >= 0; i-- { if cur.children[w[i]-'a'] == nil { cur.children[w[i]-'a'] = new(trie) } cur = cur.children[w[i]-'a'] } } return dfs(root, 1) } func dfs(cur *trie, l int) int { isLeaf, ans := true, 0 for i := 0; i < 26; i++ { if cur.children[i] != nil { isLeaf = false ans += dfs(cur.children[i], l+1) } } if isLeaf { ans += l } return ans }

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