Short Encoding of Words — LeetCode 820 Python Solution

The problem

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].

Example

Input

words = ["time", "me", "bell"]

Output

10

Explanation

A valid encoding would be s = "time#bell#" and indices = [0, 2, 5].

Python solution

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

Complexity

Pattern: Trie

Store a set of words by their shared prefixes so lookups cost the length of the word. LeetCode 820. Short Encoding of Words is filed here because LeetCode tags it Trie, which is the vocabulary this hub collects.

The trie guide has the Python template for the pattern and the 49 LeetCode problems that use it.

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