953. Verifying an Alien Dictionary

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.

Example 1:

• Input: words = [“hello”,”leetcode”], order = “hlabcdefgijkmnopqrstuvwxyz”
• Output: true
• Explanation: As ‘h’ comes before ‘l’ in this language, then the sequence is sorted.

Example 2:

• Input: words = [“word”,”world”,”row”], order = “worldabcefghijkmnpqstuvxyz”
• Output: false
• Explanation: As ‘d’ comes after ‘l’ in this language, then words[0] > words[1], hence the sequence is unsorted.

Example 3:

• Input: words = [“apple”,”app”], order = “abcdefghijklmnopqrstuvwxyz”
• Output: false
• Explanation: The first three characters “app” match, and the second string is shorter (in size.) According to lexicographical rules “apple” > “app”, because ‘l’ > ‘∅’, where ‘∅’ is defined as the blank character which is less than any other character (More info).

Constraints:

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

Submitted Code

class Solution:
    def isAlienSorted(self, words: List[str], order: str) -> bool:
        order_map = {c: i for i, c in enumerate(order)}

        for i in range(1, len(words)):
            prev_w, curr_w = words[i-1], words[i]

            for j in range(min(len(prev_w), len(curr_w))):
                if prev_w[j] != curr_w[j]:
                    if order_map[prev_w[j]] > order_map[curr_w[j]]:
                        return False  # 정렬이 틀린 경우
                    break             # 정렬이 올바른 경우
            else:   # for 루프가 break 없이 정상 종료된 경우(두 단어의 길이가 다름)
                if len(prev_w) > len(curr_w):
                    return False

        return True

Runtime: 0 ms | Beats 100.00%
Memory: 17.73 MB | Beats 74.01%

Other Solutions

1st

class Solution:
    def isAlienSorted(self, words, order):
        ind = {c: i for i, c in enumerate(order)}
        for a, b in zip(words, words[1:]):
            if len(a) > len(b) and a[:len(b)] == b:
                return False
            for s1, s2 in zip(a, b):
                if ind[s1] < ind[s2]:
                    break
                elif ind[s1] > ind[s2]:
                    return False
        return True

time complexity: 𝑂(𝑛𝑚)
space complexity: 𝑂(1)

zip()으로 인접한 두 원소끼리 묶으면 더 짧은 쪽 길이까지만 맞춰서 비교한다.