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 'a','i','u' are not present
Input : ABeeIghiObhkUul
Output : Accepted
All vowels are present
Python Program to Accept the Strings Which Contains all Vowels
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.
# 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
Time complexity : O(n)
Auxiliary Space : O(n)
Strings Which Contains all Vowels Using builtin
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:
# 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
Accept the Strings Which Contains all Vowels Using Regular Expressions
Compile a regular expression using compile() for “character is not a, e, i, o and u”. Use re.findall() to fetch the strings satisfying the above regular expression. Print output based on the result.
#import library import re
sampleInput = "aeioAEiuioea"
# regular expression to find the strings # which have characters other than a,e,i,o and u c = re. compile ( '[^aeiouAEIOU]' )
# use findall() to get the list of strings # that have characters other than a,e,i,o and u. if ( len (c.findall(sampleInput))):
print ( "Not Accepted" ) # if length of list > 0 then it is not accepted
else :
print ( "Accepted" ) # if length of list = 0 then it is accepted
|
Accepted
Accept the Strings Which Contains all Vowelsusing data structures
# Python | Program to accept the strings which contains all vowels def all_vowels(str_value):
new_list = [char for char in str_value.lower() if char in 'aeiou' ]
if new_list:
dic, lst = {}, []
for char in new_list:
dic[ 'a' ] = new_list.count( 'a' )
dic[ 'e' ] = new_list.count( 'e' )
dic[ 'i' ] = new_list.count( 'i' )
dic[ 'o' ] = new_list.count( 'o' )
dic[ 'u' ] = new_list.count( 'u' )
for i, j in dic.items():
if j = = 0 :
lst.append(i)
if lst:
return f "All vowels except {','.join(lst)} are not present"
else :
return 'All vowels are present'
else :
return "No vowels present"
# function-call str_value = "geeksforgeeks"
print (all_vowels(str_value))
str_value = "ABeeIghiObhkUul"
print (all_vowels(str_value))
|
All vowels except a,i,u are not present All vowels are present
Accept the Strings Which Contains all Vowels using set methods
The issubset() attribute of a set in Python3 checks if all the elements of a given set are present in another set.
# Python program to accept the strings # which contains all the vowels # Function for check if string # is accepted or not def check(string) :
# Checking if "aeiou" is a subset of the set of all letters in the string
if set ( "aeiou" ).issubset( set (string.lower())):
return "Accepted"
return "Not accepted"
# Driver code if __name__ = = "__main__" :
string = "SEEquoiaL"
# calling function
print (check(string))
|
Accepted
Accept the Strings Which Contains all Vowels Using collections
One approach using the collections module could be to use a Counter object to count the occurrences of each character in the string. Then, we can check if the count for each vowel is greater than 0. If it is, we can add 1 to a counter variable. At the end, we can check if the counter variable is equal to the number of vowels. If it is, the string is accepted, otherwise it is not accepted.
Here is an example of this approach:
import collections
def check(string):
# create a Counter object to count the occurrences of each character
counter = collections.Counter(string.lower())
# set of vowels
vowels = set ( "aeiou" )
# counter for the number of vowels present
vowel_count = 0
# check if each vowel is present in the string
for vowel in vowels:
if counter[vowel] > 0 :
vowel_count + = 1
# check if all vowels are present
if vowel_count = = len (vowels):
print ( "Accepted" )
else :
print ( "Not Accepted" )
# test the function string = "SEEquoiaL"
check(string) |
Accepted
Accept the Strings Which Contains all Vowels using all () method
# Python program to accept the strings # which contains all the vowels # Function for check if string # is accepted or not #using all() method def check(string):
vowels = "aeiou" #storing vowels
if all (vowel in string.lower() for vowel in vowels):
return "Accepted"
return "Not accepted"
#initializing string string = "SEEquoiaL"
# test the function print (check(string))
#this code contributed by tvsk |
Accepted
The Time Complexity of the function is O(n), and the Space Complexity is O (1) which makes this function efficient and suitable for most use cases.
Accept the Strings Which Contains all Vowels using the set difference() method
#function definition def check(s):
A = { 'a' , 'e' , 'i' , 'o' , 'u' }
#using the set difference
if len (A.difference( set (s.lower()))) = = 0 :
print ( 'accepted' )
else :
print ( 'not accepted' )
#input s = 'SEEquoiaL'
#function call check(s) |
accepted
Time complexity:O(n)
Auxiliary Space:O(n)