953. Verifying an Alien Dictionary

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.

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.

Constraints:

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

Python Solution

class Solution:
    def isAlienSorted(self, words: List[str], order: str) -> bool:
        # Create a mapping of alien alphabet to normal alphabet
        alien_to_normal = {c: i for i, c in enumerate(order)}
        
        # Compare adjacent words
        for i in range(len(words) - 1):
            word1, word2 = words[i], words[i + 1]
            
            # Compare characters until one word ends
            for j in range(min(len(word1), len(word2))):
                if word1[j] != word2[j]:
                    if alien_to_normal[word1[j]] > alien_to_normal[word2[j]]:
                        return False
                    break
            else:
                # If we get here, one word is prefix of another
                if len(word1) > len(word2):
                    return False
        
        return True

Java Solution

class Solution {
    public boolean isAlienSorted(String[] words, String order) {
        int[] alienToNormal = new int[26];
        for (int i = 0; i < order.length(); i++) {
            alienToNormal[order.charAt(i) - 'a'] = i;
        }
        
        for (int i = 0; i < words.length - 1; i++) {
            String word1 = words[i];
            String word2 = words[i + 1];
            
            for (int j = 0; j < Math.min(word1.length(), word2.length()); j++) {
                if (word1.charAt(j) != word2.charAt(j)) {
                    if (alienToNormal[word1.charAt(j) - 'a'] > alienToNormal[word2.charAt(j) - 'a']) {
                        return false;
                    }
                    break;
                }
            }
            
            // If we get here, one word is prefix of another
            if (word1.length() > word2.length()) {
                return false;
            }
        }
        
        return true;
    }
}

C++ Solution

class Solution {
public:
    bool isAlienSorted(vector& words, string order) {
        vector alienToNormal(26);
        for (int i = 0; i < order.length(); i++) {
            alienToNormal[order[i] - 'a'] = i;
        }
        
        for (int i = 0; i < words.size() - 1; i++) {
            string& word1 = words[i];
            string& word2 = words[i + 1];
            
            for (int j = 0; j < min(word1.length(), word2.length()); j++) {
                if (word1[j] != word2[j]) {
                    if (alienToNormal[word1[j] - 'a'] > alienToNormal[word2[j] - 'a']) {
                        return false;
                    }
                    break;
                }
            }
            
            // If we get here, one word is prefix of another
            if (word1.length() > word2.length()) {
                return false;
            }
        }
        
        return true;
    }
};

JavaScript Solution

/**
 * @param {string[]} words
 * @param {string} order
 * @return {boolean}
 */
var isAlienSorted = function(words, order) {
    const alienToNormal = new Array(26);
    for (let i = 0; i < order.length; i++) {
        alienToNormal[order.charCodeAt(i) - 97] = i;
    }
    
    for (let i = 0; i < words.length - 1; i++) {
        const word1 = words[i];
        const word2 = words[i + 1];
        
        for (let j = 0; j < Math.min(word1.length, word2.length); j++) {
            if (word1[j] !== word2[j]) {
                if (alienToNormal[word1.charCodeAt(j) - 97] > alienToNormal[word2.charCodeAt(j) - 97]) {
                    return false;
                }
                break;
            }
        }
        
        // If we get here, one word is prefix of another
        if (word1.length > word2.length) {
            return false;
        }
    }
    
    return true;
};

C# Solution

public class Solution {
    public bool IsAlienSorted(string[] words, string order) {
        int[] alienToNormal = new int[26];
        for (int i = 0; i < order.Length; i++) {
            alienToNormal[order[i] - 'a'] = i;
        }
        
        for (int i = 0; i < words.Length - 1; i++) {
            string word1 = words[i];
            string word2 = words[i + 1];
            
            for (int j = 0; j < Math.Min(word1.Length, word2.Length); j++) {
                if (word1[j] != word2[j]) {
                    if (alienToNormal[word1[j] - 'a'] > alienToNormal[word2[j] - 'a']) {
                        return false;
                    }
                    break;
                }
            }
            
            // If we get here, one word is prefix of another
            if (word1.Length > word2.Length) {
                return false;
            }
        }
        
        return true;
    }
}