# 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
}