Check if words are sorted according to new order of alphabets
Given a sequence of Words and the Order of the alphabet. The order of the alphabet is some permutation of lowercase letters. The task is to check whether the given words are sorted lexicographically according to order of alphabets. Return “True” if it is, otherwise “False”.
Examples:
Input : Words = [“hello”, “leetcode”], Order = “habcldefgijkmnopqrstuvwxyz”
Output : true
Input : Words = [“word”, “world”, “row”], Order = “abcworldefghijkmnpqstuvxyz”
Output : false
Explanation : As ‘d’ comes after ‘l’ in Order, thus words[0] > words[1], hence the sequence is unsorted.
Approach : The words are sorted lexicographically if and only if adjacent words are sorted. This is because order is transitive i:e if a <= b and b <= c, implies a <= c.
So our goal it to check whether all adjacent words a and b have a <= b.
To check whether for two adjacent words a and b, a <= b holds we can find their first difference. For example, "seen" and "scene" have a first difference of e and c. After this, we compare these characters to the index in order.
We have to deal with the blank character effectively. If for example, we are comparing "add" to "addition", this is a first difference of (NULL) vs "i".
Below is the implementation of above approach :
Python3
# Function to check whether Words are sorted in given Order def isAlienSorted(Words, Order): Order_index = {c: i for i, c in enumerate(Order)} for i in range(len(Words) - 1): word1 = Words[i] word2 = Words[i + 1] # Find the first difference word1[k] != word2[k]. for k in range(min(len(word1), len(word2))): # If they compare false then it's not sorted. if word1[k] != word2[k]: if Order_index[word1[k]] > Order_index[word2[k]]: return False break else: # If we didn't find a first difference, the # Words are like ("add", "addition"). if len(word1) > len(word2): return False return True # Program Code Words = ["hello", "leetcode"] Order = "habcldefgijkmnopqrstuvwxyz" # Function call to print required answer print(isAlienSorted(Words, Order)) |
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Output:
True
Time Complexity: O(N), where N is the total number of characters in all words.
Auxiliary Space: O(1)
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