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.

Constraints:

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

Examples

Example 1:
Input: words = ["hello","leetcode"], order = "hlabcdefgijkmnopqrstuvwxyz"
Output: true

Example 2:
Input: words = ["word","world","row"], order = "worldabcefghijkmnpqstuvxyz"
Output: false

Example 3:
Input: words = ["apple","app"], order = "abcdefghijklmnopqrstuvwxyz"
Output: false

How to Solve

This problem is a simple string traversal of a pair. Take two words pair one by one. Find the index of the first mismatch characters. The given character order matters. The previous word’s mismatch character index should be smaller in the given order string to be valid. The edge case is: one is a substring of another. In this case, the previous word’s length should be shorter.

If all conditions meet, the dictionary is valid.

Solution

• C++
• Java
• JavaScript
• Python
• Ruby

• /**
   * @param {string[]} words
   * @param {string} order
   * @return {boolean}
   */
  var isAlienSorted = function(words, order) {
      let prev = words.shift()
      for (let i = 0; i < words.length; ++i) {
          cur = words[i]
          if (compareTwoWords(prev, cur, order) < 0) return false
          prev = cur
      }
      return true
  };
  
  const compareTwoWords = (prev, cur, order) => {
      let idx = 0
      while (idx < Math.min(prev.length, cur.length)) {
          if (prev.charAt(idx) !== cur.charAt(idx)) {
              break
          }
          if (idx === Math.min(prev.length, cur.length) - 1) {
              return cur.length - prev.length
          }
          idx++
      }
      return order.indexOf(cur[idx]) - order.indexOf(prev[idx])
  }
  

• class VerifyingAnAlianDictionary:
      def isAlienSorted(self, words: List[str], order: str) -> bool:
          def helper(prev, cur):
              for i in range(min(len(prev), len(cur))):
                  if prev[i] != cur[i]:
                      break
                  if i == min(len(prev), len(cur)) - 1:
                      return len(cur) - len(prev)
              return order.index(cur[i]) - order.index(prev[i])
          prev = words[0]
          for cur in words[1:]:
              if helper(prev, cur) < 0:
                  return False
              prev = cur
          return True
  

• # @param {String[]} words
  # @param {String} order
  # @return {Boolean}
  def is_alien_sorted(words, order)
    prev = words.shift
    words.each do |cur|
      if compare_two_words(prev, cur, order) < 0
        return false
      end
      prev = cur
    end
    true
  end
  
  def compare_two_words(prev, cur, order)
    idx = 0
    while idx < [prev.size, cur.size].min
      break if prev[idx] != cur[idx]
      if idx == [prev.size, cur.size].min - 1
        return cur.size - prev.size
      end
      idx += 1
    end
    order.index(cur[idx]) - order.index(prev[idx])
  end
  

Complexities

• Time: O(n) – n is a total number of characters in words
• Space: O(1)