953

953. Verifying an Alien Dictionary (Easy)

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. 1 <= words.length <= 100
2. 1 <= words[i].length <= 20
3. order.length == 26
4. All characters in words[i] and order are english lowercase letters.

Companies:
Facebook

Related Topics:
Hash Table

Solution 1.

If we can change the original arary, we can map the strings in the words array to new strings using the mapping specified in order, then just check if the words are in ascending order.

// OJ: https://leetcode.com/problems/verifying-an-alien-dictionary/
// Author: github.com/lzl124631x
// Time: O(C) where C is the length of all the content in `words`
// Space: O(1)
class Solution {
public:
    bool isAlienSorted(vector<string>& words, string order) {
        char m[26];
        for (int i = 0; i < 26; ++i) m[order[i] - 'a'] = 'a' + i;
        for (auto &s : words) {
            for (char &c : s) c = m[c - 'a'];
        }
        for (int i = 1; i < words.size(); ++i) {
            if (words[i - 1] > words[i]) return false;
        }
        return true;
    }
};

Solution 2.

If we can't change the original array, we check whether each consecutive word pairs obeys the order.

// OJ: https://leetcode.com/problems/verifying-an-alien-dictionary/
// Author: github.com/lzl124631x
// Time: O(C) where C is the length of all the content in `words`.
// Space: O(1)
class Solution {
public:
    bool isAlienSorted(vector<string>& A, string order) {
        int priority[26] = {};
        for (int i = 0; i < 26; ++i) priority[order[i] - 'a'] = i;
        int N = A.size();
        for (int i = 1; i < N; ++i) {
            int j = 0, M = min(A[i - 1].size(), A[i].size());
            for (; j < M && A[i - 1][j] == A[i][j]; ++j);
            if ((j == M && A[i - 1].size() > A[i].size())
                || (j < M && priority[A[i - 1][j] - 'a'] > priority[A[i][j] - 'a'])) return false;
        }
        return true;
    }
};