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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

Solution 1

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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 c in w[::-1]:
                idx = ord(c) - 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
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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;
    }
}
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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