820-short-encoding-of-words

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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.

Solution(Python)

class Solution:
    def minimumLengthEncoding(self, words: List[str]) -> int:
        return self.triesolution(words)

    # Time Complexity: O(Σw^2)
    # Space complexity: O(Σw)
    def bruteforce(self, words: List[str]) -> int:
        good = set(words)
        for word in words:
            for k in range(1, len(word)):
                good.discard(word[k:])
        return sum(len(word) + 1 for word in good)

    # Time Complexity: O(Σw)
    # Space complexity: O(Σw)
    def triesolution(self, words: List[str]) -> int:
        words = list(set(words))  # remove duplicates

        def Trie():
            return collections.defaultdict(Trie)

        trie = Trie()

        nodes = [reduce(dict.__getitem__, word[::-1], trie) for word in words]
        return sum(len(word) + 1 for i, word in enumerate(words) if len(nodes[i]) == 0)