# [Leet Code] Count the Number of Consistent Strings

Problem:

You are given a string `allowed` consisting of distinct characters and an array of strings `words`. A string is consistent if all characters in the string appear in the string `allowed`.

Return the number of consistent strings in the array `words`.

Example 1:

`Input: allowed = "ab", words = ["ad","bd","aaab","baa","badab"]Output: 2Explanation: Strings "aaab" and "baa" are consistent since they only contain characters 'a' and 'b'.`

Example 2:

`Input: allowed = "abc", words = ["a","b","c","ab","ac","bc","abc"]Output: 7Explanation: All strings are consistent.`

Example 3:

`Input: allowed = "cad", words = ["cc","acd","b","ba","bac","bad","ac","d"]Output: 4Explanation: Strings "cc", "acd", "ac", and "d" are consistent.`

Constraints:

• `1 <= allowed.length <= 26`
• `1 <= words[i].length <= 10`
• The characters in `allowed` are distinct.
• `words[i]` and `allowed` contain only lowercase English letters.

Solution:

`class Solution(object):    def countConsistentStrings(self, allowed, words):        """        :type allowed: str        :type words: List[str]        :rtype: int        """        counter = 0        for word in words:            stop = 0            for letter in word:                if letter not in allowed:                    stop += 1            if stop == 0:                counter += 1                        return counter`

Explanation:

we create our counter, then we cyce through the words, and then again through each of the letters in the word. We check if there are any letters that are not in the “allowed” letters, and if so, we increment a stop variable. If stop != 0, that means that there were some letters that weren’t in the allowed variable, so we continue on, otherwise, we increment the counter variable if stop == 0.

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