Python – Retain first N Elements of a String and Replace the Remaining by K

Given a String, retain first N elements and replace rest by K.

Input : test_str = ‘neveropen’, N = 5, K = “@” 
Output :neveropen@@@@@@@@ 
Explanation : First N elements retained and rest replaced by K.

Input : test_str = ‘neveropen’, N = 5, K = “*” 
Output :neveropen******** 
Explanation : First N elements retained and rest replaced by K.

Method #1 : Using * operator + len() + slicing

In this, slicing is used to retain N, and then the length of remaining is extracted by subtracting the total length extracted by len(), from N, and then repeat K char using * operator.

The original string is :neveropen
The resultant string : geek@@@@@@@@@

Time Complexity: O(n) -> string slicing

Auxiliary Space: O(n)

Method #2 : Using ljust() + slicing + len()

In this, the task of assigning remaining characters is done using ljust, rather than the * operator.

The original string is :neveropen
The resultant string : geek@@@@@@@@@

Time Complexity: O(n) -> string slicing

Auxiliary Space: O(n)

Method 3 (Using replace() method):

Use the replace method to replace the characters except the first N elements.

The original string is :neveropen
The resultant string is : geek@@@@@@@@@

Time Complexity: O(n) -> average of string slicing and replace function

Auxiliary Space: O(n)

Method 4: Using a loop and concatenation.

Algorithm:

1. Initialize an empty string, say res.
2. Traverse through the characters of the input string, test_str.
3. Check if the index is less than N.
   a. If the index is less than N, append the character at that index to res.
   b. Else, append the character K to res.
4. Return the resultant string, res.

The original string is :neveropen
The resultant string : geek@@@@@@@@@

Time Complexity: O(n), where n is the length of the input string. The loop runs n times.

Auxiliary Space Complexity: O(n), where n is the length of the input string. The size of the resultant string can be the same as the size of the input string, in case N is less than or equal to the length of the input string.

Method 5: Using string format(): 

Algorithm:

1.Initialize the string to be modified and the length needed and the character to replace remaining characters with.
2.Use string slicing to obtain the first N characters of the original string and concatenate it with K multiplied by the difference between the length of the original string and N.

The original string is :neveropen
The resultant string : geek@@@@@@@@@

Time Complexity: O(N), where N is the length of the original string. The slicing operation takes O(N) time.

Auxiliary Space: O(N), where N is the length of the original string. The resultant string can take up to N characters.

Method 6: Using map() function and lambda function

Steps were to solve using map and lambda:

• Initialize the original string test_str, number of characters to retain N and the character to replace the remaining characters with K.
• Concatenate the first N characters of the original string with K multiplied by the difference between the length of the original string and N.
• Print the original string and the resultant string.

The original string is :neveropen
The resultant string : geek@@@@@@@@@

Time Complexity:
The time complexity of this code is O(N) because the number of iterations in the for loop is proportional to the value of N.

Space Complexity:
The space complexity of this code is O(N) because we are storing the output string in a new variable which has a length equal to the length of the original string.

Method 7: Using string concatenation and conditional operator

The idea is to iterate over the original string and concatenate the first N characters to the result string. For the remaining characters, replace them with the specified character K.

The resultant string : geek@@@@@@@@@

Time complexity:  O(n), where n is the length of the input string. 
Auxiliary space: O(n) as we are using an additional string to store the result.