Given a string, the task is to check if every vowel is present or not. We consider a vowel to be present if it is present in upper case or lower case. i.e. ‘a’, ‘e’, ‘i’.’o’, ‘u’ or ‘A’, ‘E’, ‘I’, ‘O’, ‘U’ .
Examples :
Input : geeksforgeeks Output : Not Accepted All vowels except 'o' are not present Input : ABeeIghiObhkUul Output : Accepted All vowels are present
Approach : Firstly, create set of vowels using set() function. Check for each character of the string is vowel or not, if vowel then add into the set s. After coming out of the loop, check length of the set s, if length of set s is equal to the length of the vowels set then string is accepted otherwise not.
Below is the implementation :
Python3
# Python program to accept the strings # which contains all the vowels # Function for check if string # is accepted or not def check(string) : string = string.lower() # set() function convert "aeiou" # string into set of characters # i.e.vowels = {'a', 'e', 'i', 'o', 'u'} vowels = set ( "aeiou" ) # set() function convert empty # dictionary into empty set s = set ({}) # looping through each # character of the string for char in string : # Check for the character is present inside # the vowels set or not. If present, then # add into the set s by using add method if char in vowels : s.add(char) else : pass # check the length of set s equal to length # of vowels set or not. If equal, string is # accepted otherwise not if len (s) = = len (vowels) : print ( "Accepted" ) else : print ( "Not Accepted" ) # Driver code if __name__ = = "__main__" : string = "SEEquoiaL" # calling function check(string) |
Accepted
Alternate Implementation :
Python3
def check(string): string = string.replace( ' ' , '') string = string.lower() vowel = [string.count( 'a' ), string.count( 'e' ), string.count( 'i' ), string.count( 'o' ), string.count( 'u' )] # If 0 is present int vowel count array if vowel.count( 0 ) > 0 : return ( 'not accepted' ) else : return ( 'accepted' ) # Driver code if __name__ = = "__main__" : string = "SEEquoiaL" print (check(string)) |
accepted
Alternate Implementation 2.0 :
Python3
# Python program for the above approach def check(string): if len ( set (string.lower()).intersection( "aeiou" )) > = 5 : return ( 'accepted' ) else : return ( "not accepted" ) # Driver code if __name__ = = "__main__" : string = "geeksforgeeks" print (check(string)) |
not accepted
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