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Python | Program to accept the strings which contains all vowels

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  • Difficulty Level : Easy
  • Last Updated : 30 Jan, 2023
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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

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)


Output

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))


Output

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))


Output

not accepted

Alternate Implementation 3.0 (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.

Python3




#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


Output

Accepted

Alternate Implementation 4.0 (using data structures):

Python3




# 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))
 
# contribute by saikot


Output

All vowels except a,i,u are not present
All vowels are present

Alternate Approach (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.

Python3




# 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))


Output

Accepted

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:

Python3




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)


Output

Accepted

The time complexity of the above code is O(n), where n is the length of the input string. This is because the code iterates through each character in the input string and performs a constant number of operations on each character.

The auxiliary space complexity of the above code is also O(n), because the code creates two sets, one to store the vowels and one to store the vowels found in the input string, both of which have a size that is equal to the length of the input string. In addition to these sets, the code also creates a list to store the count of each vowel in the input string, which also has a size equal to the length of the input string. Overall, the auxiliary space complexity is O(n).

Alternate Approach (using  all () method) 

Python3




# 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"
 
 
 
#intitalizing string
string = "SEEquoiaL"
# test the function
print(check(string))
 
#this code contributed by tvsk


Output

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.

Approach: using the set difference() method 

Python3




#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)


Output

accepted

Time complexity:O(n)
Auxiliary Space:O(n)


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