Given a String, perform split on vowels.
Example:
Input : test_str = ‘GFGaBst’
Output : [‘GFG’, ‘Bst’]
Explanation : a is vowel and split happens on that.Input : test_str = ‘GFGaBstuforigeeks’
Output : [‘GFG’, ‘Bst’, ‘f’, ‘r’, ‘g’, ‘ks’]
Explanation : a, e, o, u, i are vowels and split happens on that.
Naive approach:
- Initialize variable vowels to contain all the vowels.
- Initialize an empty list result and a variable temp to an empty string.
- Iterate through each character in the input string test_str.
- For each character, check if it is a vowel (by checking if it is in the vowels variable).
- If the character is a vowel and the temp variable is not empty, append temp to the result list and reset temp to an empty string.
- If the character is not a vowel, add it to the temp variable.
- After the iteration, if the temp variable is not empty, append it to the result list.
- Return the result list.
def split_on_vowels(test_str):
vowels = 'aeiouAEIOU'
result = []
temp = ""
for char in test_str:
if char in vowels:
if temp ! = "":
result.append(temp)
temp = ""
else :
temp + = char
if temp ! = "":
result.append(temp)
return result
test_str = 'GFGaBstuforigeeks'
print (split_on_vowels(test_str))
|
['GFG', 'Bst', 'f', 'r', 'g', 'ks']
Time Complexity: O(N)
Auxiliary Space: O(N)
Method 1 : Using regex() + split()
In this, we use regex split() which accepts multiple characters to perform split, passing list of vowels, performs split operation over string.
# Python3 code to demonstrate working of # Split String on vowels # Using split() + regex import re
# initializing strings test_str = 'GFGaBste4oCS'
# printing original string print ( "The original string is : " + str (test_str))
# splitting on vowels # constructing vowels list # and separating using | operator res = re.split( 'a|e|i|o|u' , test_str)
# printing result print ( "The splitted string : " + str (res))
|
The original string is : GFGaBste4oCS The splitted string : ['GFG', 'Bst', '4', 'CS']
Time Complexity: O(n), where n is the length of the string “test_str”. The “re.split” function splits the string by searching for specified characters (vowels), which takes linear time proportional to the length of the string.
Auxiliary space: O(1), as it uses a constant amount of memory regardless of the size of the input string “test_str”.
Method 2 : Using replace() and split().
First replace all vowels in string with “*” and then split the string by “*” as delimiter
# Python3 code to demonstrate working of # Split String on vowels # initializing strings test_str = 'GFGaBste4oCS'
# printing original string print ( "The original string is : " + str (test_str))
# splitting on vowels vow = "aeiouAEIOU"
for i in test_str:
if i in vow:
test_str = test_str.replace(i, "*" )
res = test_str.split( "*" )
# printing result print ( "The splitted string : " + str (res))
|
The original string is : GFGaBste4oCS The splitted string : ['GFG', 'Bst', '4', 'CS']
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #3 : Using replace(),split() and ord() methods
# Python3 code to demonstrate working of # Split String on vowels # initializing strings test_str = 'GFGaBste4oCS'
# printing original string print ( "The original string is : " + str (test_str))
# splitting on vowels x = [ 97 , 101 , 105 , 111 , 117 , 65 , 69 , 73 , 79 , 85 ]
for i in test_str:
if ord (i) in x:
test_str = test_str.replace(i, "*" )
res = test_str.split( '*' )
# printing result print ( "The splitted string : " + str (res))
|
The original string is : GFGaBste4oCS The splitted string : ['GFG', 'Bst', '4', 'CS']
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #4 : Using operator.countOf() method
import operator as op
def split_on_vowels(test_str):
vowels = 'aeiouAEIOU'
result = []
temp = ""
for char in test_str:
if op.countOf(vowels, char) > 0 :
if temp ! = "":
result.append(temp)
temp = ""
else :
temp + = char
if temp ! = "":
result.append(temp)
return result
test_str = 'GFGaBstuforigeeks'
print (split_on_vowels(test_str))
|
['GFG', 'Bst', 'f', 'r', 'g', 'ks']
Time Complexity: O(n)
Auxiliary Space: O(n)
Method5# :using the ‘itertools.groupby ‘function from the ‘itertools’ module
import itertools
# Define the string to be split test_str = 'GFGaBste4oCS'
# Define the vowels to split the string on vowels = 'aeiouAEIOU'
# Print the original string print ( "The original string is:" , test_str)
# Use itertools.groupby to group adjacent characters in test_str based on if they are in vowels res = [ list (g) for k, g in itertools.groupby(test_str, key = lambda x: x not in vowels) if k]
# Join each group of characters into a string res = [''.join(substring) for substring in res]
# Print the final split string print ( "The split string is:" , res)
#this code is contributed by Asif_shaik |
The original string is: GFGaBste4oCS The split string is: ['GFG', 'Bst', '4', 'CS']
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #6: Using translate method
Algorithm:
- Initialize a string test_str.
- Create a translation table that replaces vowels with spaces using the str.maketrans() method.
- Apply the translation table to the test_str using the translate() method and store it in a variable named trans_str.
- Split the trans_str on spaces and store the result in a list named res.
- Print the result.
# initializing string test_str = 'GFGaBste4oCS'
# create a translation table that replaces vowels with spaces trans_table = str .maketrans( 'aeiouAEIOU' , ' ' * 10 )
# split the string on spaces res = test_str.translate(trans_table).split()
# printing result print ( "The splitted string : " + str (res))
#This code is contributed by Vinay Pinjala. |
The splitted string : ['GFG', 'Bst', '4', 'CS']
Time complexity: The time complexity of this code is O(n) because the maketrans(), translate() and split() methods take linear time.
Auxiliary Space: The space complexity of this code is O(n) because we are creating a new string and a new list to store the results.