0953. Verifying an Alien Dictionary
https://leetcode.com/problems/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](https://en.wikipedia.org/wiki/Lexicographical_order)).
Constraints:
1 <= words.length <= 100
1 <= words[i].length <= 20
order.length == 26
All characters in
words[i]
andorder
are English lowercase letters.
ac
class Solution {
public boolean isAlienSorted(String[] words, String order) {
// Edge cases
if (words.length <= 1) {
return true;
}
int[] letterOrder = new int[26];
for (int i = 0; i < order.length(); i++) {
char c = order.charAt(i);
letterOrder[c-'a'] = i;
}
for (int i = 1; i < words.length; i++) {
if (!isCorrectOrder(words[i-1], words[i], letterOrder)) return false;
}
return true;
}
private boolean isCorrectOrder(String word1, String word2, int[] letterOrder) {
int len1 = word1.length();
int len2 = word2.length();
for (int i = 0; i < len1 && i < len2; i++) {
if (word1.charAt(i) == word2.charAt(i)) continue;
// Meet different char
int order1 = letterOrder[word1.charAt(i) - 'a'];
int order2 = letterOrder[word2.charAt(i) - 'a'];
return order1 < order2;
}
// 1) 2 words are the same; 2) word1 is prefix of word2, e.g. [app, apple]
return len1 <= len2;
}
}
// Key: how to compare order in a array - just compare every 2 items
// O(NS) time: N - length of words array, S - average length of word. O(1) space.
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