Verifying an Alien Dictionary - Hash Table - Easy - LeetCode
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Verifying an Alien Dictionary - Hash Table - Easy - LeetCode

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  • 1The alien language uses a permutation of English lowercase letters for its alphabet order.
  • 2To verify if words are sorted, a mapping of characters to their indices in the alien order is created.
  • 3The algorithm checks each pair of words lexicographically based on the alien alphabet, returning true if sorted.

AI-generated summary · May not capture all nuances

Key Insight
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"The alien language uses a permutation of English lowercase letters for its alphabet order."

Verifying an Alien Dictionary - Hash Table - Easy - LeetCode

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).

Constraints:

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

public class Solution {
 public bool IsAlienSorted(string[] words, string order) {
 var map = new Dictionary<char,int>();
 for(int i=0;i<order.Length;i++){
 map.Add(order[i],i);
 }
 
 if(words.Length==1){
 return true;
 }
 
 for(int i=1;i<words.Length;i++){
 if(!Helper(map,words[i-1],words[i],0)){
 return false;
 }
 }
 
 return true;
 }
 
 private bool Helper(Dictionary<char,int> map, string word1, string word2, int index){ 
 
 if(index>=word1.Length || index>=word2.Length){
 return word2.Length>word1.Length;
 }
 
 if(map[word1[index]]<map[word2[index]]){ 
 return true;
 }
 
 if(map[word1[index]]>map[word2[index]]){ 
 return false;
 }
 
 return Helper(map,word1,word2,index+1);
 }
}

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Where m and n are the number of strings and length of strings.

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sumitc91

Published on 30 September 2020 · 1 min read · 273 words

Part of AskGif Blog · coding

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