953. Verifying an Alien Dictionary
Tags: ‘Hash Table’
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 lexicographicaly 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).
Note:
1 <= words.length <= 100
1 <= words[i].length <= 20
order.length == 26
- All characters in
words[i]
andorder
are english lowercase letters.
Solution
public boolean isAlienSorted(String[] words, String order) {
// We can also use array for hash, since char set is predefined
// int[] index = new int[26];
// for (int i = 0; i < order.length(); ++i)
// index[order.charAt(i) - 'a'] = i;
Map<Character, Integer> map = new HashMap<>();
for (int i = 0; i < order.length(); i++) {
char c = order.charAt(i);
map.put(c, i);
}
for (int i = 0; i < words.length - 1; i++) {
String before = words[i];
String after = words[i+1];
for (int j = 0; j < before.length(); j++) {
// iterate through before
// if before ends, after doesn't matter
int index1 = map.get(before.charAt(j));
if (j >= after.length()) {
return false;
}
int index2 = map.get(after.charAt(j));
if (index1 < index2) {
break; // valid. Continue to next word;
} else if (index1 > index2) {
return false; // not valid, just return
} else {
continue; // equal, continue check next char
}
}
}
return true;
}