Verifying an Alien Dictionary

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.


The words are sorted lexicographically if and only if adjacent words are. This is because order is transitive: a <= b and b <= c implies a <= c.


Let’s check whether all adjacent words a and b have a <= b.

To check whether a <= b for two adjacent words a and b, we can find their first difference. For example, "applying" and "apples" have a first difference of y vs e. After, we compare these characters to the index in order.

Care must be taken to deal with the blank character effectively. If for example, we are comparing "app" to "apply", this is a first difference of (null) vs "l".

class Solution {
    public boolean isAlienSorted(String[] words, String order) {
        Map<Character, Integer> characterOrderMap = new HashMap<>();
        for (int i = 0; i < order.length(); i++) {
            characterOrderMap.put(order.charAt(i), i);
        for (int i = 0; i < words.length - 1; i++) {
            if (!compare(words[i], words[i + 1], map)){
                return false;
        return true;
    private boolean compare(String s1, String s2, Map<Character, Integer> characterOrderMap) {
        int l1 = s1.length(), l2 = s2.length();
        for (int i = 0, j = 0; i < l1 && j < l2; i++, j++) {
            if (s1.charAt(i) != s2.charAt(j)) {
                if (map.get(s1.charAt(i)) > characterOrderMap.get(s2.charAt(j))) {
                    return false;
                } else {
                    return true;
        if (l1 > l2) return false;
        return true;

Complexity Analysis

  • Time Complexity: O(C), where C is the total content of words.
  • Space Complexity: O(1).

Categories: Map

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