# 820. Short Encoding of Words

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

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.
``````

• code
``````class Solution:
def minimumLengthEncoding(self, words: List[str]) -> int:
words = list(set(words))
m = {}

for word in words:
node = m
for c in word[::-1]:
if c not in node:
node[c] = {}
node = node[c]

res = 0
for word in words:
node = m
for c in word[::-1]:
node = node[c]
if not node: # reach to the end, otherwise like 'me' with 'time', don't need to add 'me'
res += len(word) + 1

return res

``````
• code using defaultdict
``````class Solution:
def minimumLengthEncoding(self, words: List[str]) -> int:
words = list(set(words))
TrieNode = lambda: defaultdict(TrieNode)
t = TrieNode()

for word in words:
node = t
for c in word[::-1]:
# if c not in node:
# node[c] = {}
node = node[c]

res = 0
for word in words:
node = t
for c in word[::-1]:
node = node[c]
if not node: # reach to the end, otherwise like 'me' with 'time', don't need to add 'me'
res += len(word) + 1

return res
``````