Solution.py

"""

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.


"""


class Solution:
    def minimumLengthEncoding(self, words: List[str]) -> int:
        words.sort(key=lambda x: -len(x))
        cnt = 0
        tree = {}
        for word in words:
            curr = tree
            is_new = False
            for i in range(len(word) - 1, -1, -1):
                c = word[i]
                if c not in curr:
                    curr[c] = {}
                    is_new = True
                curr = curr[c]
            if is_new:
                cnt += len(word) + 1
        return cnt