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 : neveropen
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.
Python3
# Python program to accept the strings# which contains all the vowels# Function for check if string# is accepted or notdef 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 codeif __name__ == "__main__" : string = "SEEquoiaL" # calling function check(string) |
Accepted
Time complexity : O(n)
Auxiliary Space : O(n)
Strings Which Contains all Vowels Using builtin
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 codeif __name__ == "__main__": string = "SEEquoiaL" print(check(string)) |
accepted
Alternate Implementation:
Python3
# Python program for the above approachdef check(string): if len(set(string.lower()).intersection("aeiou")) >= 5: return ('accepted') else: return ("not accepted")# Driver codeif __name__ == "__main__": string = "neveropen" 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.
Python3
#import libraryimport resampleInput = "aeioAEiuioea"# regular expression to find the strings# which have characters other than a,e,i,o and uc = 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 acceptedelse: print("Accepted") # if length of list = 0 then it is accepted |
Accepted
Accept the Strings Which Contains all Vowelsusing data structures
Python3
# Python | Program to accept the strings which contains all vowelsdef 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-callstr_value = "neveropen"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.
Python3
# Python program to accept the strings# which contains all the vowels# Function for check if string# is accepted or notdef 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 codeif __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:
Python3
import collectionsdef 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 functionstring = "SEEquoiaL"check(string) |
Accepted
Accept the Strings Which Contains all Vowels 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() methoddef check(string): vowels = "aeiou" #storing vowels if all(vowel in string.lower() for vowel in vowels): return "Accepted" return "Not accepted"#initializing stringstring = "SEEquoiaL"# test the functionprint(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
Python3
#function definitiondef 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')#inputs='SEEquoiaL'#function callcheck(s) |
accepted
Time complexity:O(n)
Auxiliary Space:O(n)
