Verifying an Alien Dictionary

Published: Sep 8, 2022

Introduction

The given dictionary letters’ order matters. Hash table is a good data structure to save the letter order. However, in case of Python, string’s index or find method can do the same, so Python doesn’t need Hash table. Other than that, this problem is a simple string traversal of a pair.

Problem Description

In an alien language, surprisingly, they also use English lowercase letters, but possibly in a different order. The order of the alphabet is some permutation of lowercase letters.

Given a sequence of words written in the alien language, and the order of the alphabet, return true if and only if the given words are sorted lexicographically in this alien language.

Constraints:

• 1 <= words.length <= 100
• 1 <= words[i].length <= 20
• order.length == 26
• All characters in words[i] and order are English lowercase letters.

https://leetcode.com/problems/verifying-an-alien-dictionary/

Examples

Example 1:
Input: words = ["hello","leetcode"], order = "hlabcdefgijkmnopqrstuvwxyz"
Output: true

Example 2:
Input: words = ["word","world","row"], order = "worldabcefghijkmnpqstuvxyz"
Output: false

Example 3:
Input: words = ["apple","app"], order = "abcdefghijklmnopqrstuvwxyz"
Output: false

Analysis

Take two words pair one by one. Find the index of the first mismatch characters. The previous word’s mismatch character index should be smaller if the string is valid. We should consider one more case. If one is a substring of another, the previous word’s length should be shorter. If the conditions don’t meet, given word list is invalid.

Solution

class VerifyingAnAlianDictionary:
    def isAlienSorted(self, words: List[str], order: str) -> bool:
        def helper(prev, cur):
            for i in range(min(len(prev), len(cur))):
                if prev[i] != cur[i]:
                    break
                if i == min(len(prev), len(cur)) - 1:
                    return len(cur) - len(prev)
            return order.index(cur[i]) - order.index(prev[i])
        prev = words[0]
        for cur in words[1:]:
            if helper(prev, cur) < 0:
                return False
            prev = cur
        return True

Complexities

• Time: O(n) – n is a total number of characters in words
• Space: O(1)