Given a string, the task is to write a Python program to replace every Nth character in a string by the given value K.
Examples:
Input : test_str = “neveropen is best for all Lazyroar”, K = ‘$’, N = 5
Output : Lazyroar$orge$ks i$ bes$ for$all $eeks
Explanation : Every 5th character is converted to $.
Input : test_str = “neveropen is best for all Lazyroar”, K = ‘*’, N = 5
Output : Lazyroar*orge*ks i* bes* for*all *eeks
Explanation : Every 5th occurrence is converted to *.
Method 1 : Using loop and enumerate()
In this, we perform an iteration of each character and check if its Nth by performing modulo, i.e finding remainder by N. If its Nth occurrence, the character is replaced by K.
Example
Python3
# initializing stringtest_str = "neveropen is best for all Lazyroar"# printing original stringprint("The original string is : " + str(test_str))# initializing KK = '$'# initializing NN = 5res = ''for idx, ele in enumerate(test_str): # add K if idx is multiple of N if idx % N == 0 and idx != 0: res = res + K else: res = res + ele# printing resultprint("String after replacement : " + str(res)) |
Output:
The original string is : neveropen is best for all Lazyroar
String after replacement : Lazyroar$orge$ks i$ bes$ for$all $eeks
The time complexity of the given code is O(N), where N is the length of the input string “test_str”.
The auxiliary space used by the code is also O(N), where N is the length of the input string “test_str”.
Method 2 : Using generator expression, join() and enumerate()
In this, the construction of string happens using join(). The enumerate(), helps to get required indices. The generator expression provides a shorthand approach to this problem.
Example
Python3
# initializing stringtest_str = "neveropen is best for all Lazyroar"# printing original stringprint("The original string is : " + str(test_str))# initializing KK = '$'# initializing NN = 5res = ''.join(ele if idx % N or idx == 0 else K for idx, ele in enumerate(test_str))# printing resultprint("String after replacement : " + str(res)) |
Output:
The original string is : neveropen is best for all Lazyroar
String after replacement : Lazyroar$orge$ks i$ bes$ for$all $eeks
The time and space complexity of all the methods are the same:
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 3 : Using lists
Python3
# initializing stringtest_str = "neveropen is best for all Lazyroar"# printing original stringprint("The original string is : " + str(test_str))# initializing KK = '$'# initializing NN = 5x=list(test_str)ns=""for i in range(0,len(x)): if(i!=0 and i%5==0): ns+=K else: ns+=test_str[i]# printing resultprint("String after replacement : " + str(ns)) |
The original string is : neveropen is best for all Lazyroar String after replacement : Lazyroar$orge$ks i$ bes$ for$all $eeks
The time complexity of this code is O(n), where n is the length of the input string test_str.
The auxiliary space used by the code is also O(n), where n is the length of the input string.
Method 4:Using string concatenation
- Initialize a string ‘test_str’ with the value “neveropen is best for all Lazyroar”.
- Print the original string.
- Initialize a character ‘K’ with the value ‘$’.
- Initialize an integer ‘N’ with the value 5.
- Create an empty string ‘new_str’ to store the modified string.
- Loop through each character of the string ‘test_str’ using the ‘range()’ function and ‘len()’ function.
- Check if the current index is a multiple of N using the modulo operator ‘%’.
- If the index is a multiple of N, add the character K to the new string ‘new_str’.
- If the index is not a multiple of N, add the original character from ‘test_str’ to the new string ‘new_str’.
- Print the final modified string ‘new_str’.
Python3
# initializing stringtest_str = "neveropen is best for all Lazyroar"# printing original stringprint("The original string is : " + str(test_str))# initializing KK = '$'# initializing NN = 5# create a new empty string to store the modified stringnew_str = ""# loop through the characters in the original stringfor i in range(len(test_str)): # check if the current index is a multiple of N if (i+1) % N == 0: # add the replacement character K to the new string new_str += K else: # add the original character to the new string new_str += test_str[i]# printing resultprint("String after replacement : " + str(new_str))#This code is contributed by Vinay Pinjala. |
The original string is : neveropen is best for all Lazyroar String after replacement : geek$forg$eks $s be$t fo$ all$geek$
The time complexity of the algorithm is O(N), where N is the length of the string ‘test_str’. This is because we are looping through each character of the string once.
The space complexity of the algorithm is also O(N), where N is the length of the string ‘test_str’. This is because we are creating a new string ‘new_str’ to store the modified string, which will have the same length as the original string in the worst case.
