Create 1684.CounttheNumberofConsistentStrings.py

Author: YuriSpiridonov

Easy/1684.CounttheNumberofConsistentStrings.py

@@ -0,0 +1,50 @@
+'''
+    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:
+    Input: allowed = "ab", words = ["ad","bd","aaab","baa","badab"]
+    Output: 2
+    Explanation: Strings "aaab" and "baa" are consistent since 
+                 they only contain characters 'a' and 'b'.
+
+    Example:
+    Input: allowed = "abc", words = ["a","b","c","ab","ac","bc","abc"]
+    Output: 7
+    Explanation: All strings are consistent.
+
+    Example:
+    Input: allowed = "cad", words = ["cc","acd","b","ba","bac","bad","ac","d"]
+    Output: 4
+    Explanation: Strings "cc", "acd", "ac", and "d" are consistent.
+
+    Constraints:
+        - 1 <= words.length <= 10^4
+        - 1 <= allowed.length <= 26
+        - 1 <= words[i].length <= 10
+        - The characters in allowed are distinct.
+        - words[i] and allowed contain only lowercase 
+          English letters.
+'''
+#Difficulty: Easy
+#74 / 74 test cases passed.
+#Runtime: 248 ms
+#Memory Usage: 16.3 MB
+
+#Runtime: 248 ms, faster than 75.00% of Python3 online submissions for Count the Number of Consistent Strings.
+#Memory Usage: 16.3 MB, less than 25.00% of Python3 online submissions for Count the Number of Consistent Strings.
+
+class Solution:
+    def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
+        count = len(words)
+        for word in words:
+            word = set(word)
+            for char in word:
+                if char not in allowed:
+                    count -= 1
+                    break
+        return count