Given a String S, a character array ch[], a number N and a replacing character, the task is to replace every Nth occurrence of each character of the character array ch[] in the string with the given replacing character.
Input: S = “GeeksforGeeks”, ch[] = {‘G’, ‘e’, ‘k’}, N = 2, replacing_character = ‘#’
Output: Ge#ksfor#ee#s
Explanation:
In the given string S, the second occurrence of the ‘G’, ‘e’, ‘K’ is replaced with ‘#’
Input: S = abcdeahu, ch[] = {‘a’, ‘d’, ‘u’}, N = 1, replacing_character = ‘#’
Output: #bc#eah#
Explanation:
In the given string S, the first occurrence of the ‘a’, ‘d’, ‘u’ is replaced with ‘#’
Method 1: Naive approach
In this approach, the general idea is to store every Nth Occurrence index of every character in another array and replacing them with the given another character.
Below is the implementation of the above approach.
# Python implementation to replace nth # occurrence of the every character # in a string # Function to replace the Nth occurrence # of the character of string def replacer(initial_string, ch,
replacing_character, occurrence):
# breaking a string into it's
# every single character in list
lst1 = list (initial_string)
lst2 = list (ch)
# List to store the indexes in which
# it is replaced with the
# replacing_character
lst3 = []
# Loop to find the Nth occurrence of
# given characters in the string
for i in lst2:
if (lst1.count(i)> = occurrence):
count = 0
for j in range ( 0 , len (initial_string)):
if (i = = lst1[j]):
count + = 1
if (count = = occurrence):
lst3.append(j)
for i in lst3:
# Replacing that particular index
# with the requested character
lst1[i] = replacing_character
print (''.join(lst1))
# Driver Code: if __name__ = = '__main__' :
initial_string = 'GeeksforGeeks'
ch = [ 'G' , 'e' , 'k' ]
occurrence = 2
replacing_character = '#'
replacer(initial_string, ch,
replacing_character, occurrence)
|
Ge#ksfor#ee#s
Time complexity: O(n^2), where n is the length of the input string.
Auxiliary space: O(n), where n is the length of the input string.
Method 2: Using find() method
In this approach, the idea is to use the find() function to find the Nth occurrence of the character in the given string S and replace it with another giver character.
Below is the implementation of the above approach.
# Python implementation to replace nth # occurrence of the every character # in a string using find() function # Function to replace the Nth occurrence # of the character of string def replacer(initial_string, ch,
replacing_character, occurrence):
# breaking a string into
# it's every single character
lst1 = list (initial_string)
lst2 = list (ch)
# Loop to find the occurrence
# of the character in the string
# and replace it with the given
# replacing_character
for i in lst2:
sub_string = i
val = - 1
for i in range ( 0 , occurrence):
val = initial_string.find(sub_string,
val + 1 )
lst1[val] = replacing_character
print (''.join(lst1))
# Driver Code: if __name__ = = '__main__' :
initial_string = 'GeeksforGeeks'
ch = [ 'G' , 'e' , 'k' ]
occurrence = 2
replacing_character = '#'
replacer(initial_string, ch,
replacing_character, occurrence)
|
Ge#ksfor#ee#s
Time complexity: O(n*m), where n is the length of the initial_string and m is the length of the list ch.
Auxiliary Space: O(n), where n is the length of the initial_string.
Method 3: Using startswith() method
In this approach, the idea is to use the startswith() function of the python to find the index of the character where the occurrence of the character is equal to the given Nth occurrence and then replace with the given replacing character.
Below is the implementation of the above approach.
# Python implementation to replace nth # occurrence of the every character # in a string # Function to replace the Nth occurrence # of the character of string def replacer(initial_string, ch,
replacing_character, occurrence):
# breaking a string into
# it's every single character
lst1 = list (initial_string)
lst2 = list (ch)
# Loop to find the occurrence
# of the character in the string
# and replace it with the given
# replacing_character
for j in lst2:
sub_string = j
checklist = [i for i in range ( 0 , len (initial_string))
if initial_string[i:].startswith(sub_string)]
if len (checklist)> = occurrence:
lst1[checklist[occurrence - 1 ]] = replacing_character
print (''.join(lst1))
# Driver Code: if __name__ = = '__main__' :
initial_string = 'GeeksforGeeks'
ch = [ 'G' , 'e' , 'k' ]
occurrence = 2
replacing_character = '#'
replacer(initial_string, ch,
replacing_character, occurrence)
|
Ge#ksfor#ee#s