Sometimes, we wish to manipulate a string in such a way in which we might need to add additional K at the end of string in case of filling the missing bits or any other specific requirement. The solution to this kind of problems is always handy and is good if one has knowledge of it. Let’s discuss certain ways in which this can be solved.
Method #1 : Using ljust() This task can be performed using the simple inbuilt string function of ljust in which we just need to pass N times required and the element to right pad, in this case being K.
Python3
# Python3 code to demonstrate # Append K character N times # using ljust() # initializing string test_string = 'GFG' # printing original string print ( "The original string : " + str (test_string)) # initializing K K = 'M' # No. of K required N = 5 # using ljust() # Append K character N times res = test_string.ljust(N + len (test_string), K) # print result print ( "The string after adding trailing K : " + str (res)) |
The original string : GFG The string after adding trailing K : GFGMMMMM
Method #2 : Using format() String formatting using the format function can be used to perform this task easily, we just mention the number of elements total, element needed to pad, and direction of padding, in this case right.
Python3
# Python3 code to demonstrate # Append K character N times # using format() # initializing string test_string = 'GFG' # printing original string print ( "The original string : " + str (test_string)) # initializing K K = '0' # No. of zeros required N = 5 # using format() # Append K character N times temp = '{:<' + K + str ( len (test_string) + N) + '}' res = temp. format (test_string) # print result print ( "The string after adding trailing K : " + str (res)) |
The original string : GFG The string after adding trailing K : GFG00000
Method #3: Using +,* operators
Python3
# Python3 code to demonstrate # Append K character N times # initializing string test_string = 'GFG' # printing original string print ( "The original string : " + str (test_string)) # initializing K K = 'M' # No. of K required N = 5 # Append K character N times test_string = test_string + K * N # print result print ( "The string after adding trailing K : " + str (test_string)) |
The original string : GFG The string after adding trailing K : GFGMMMMM
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #4: Using repeat()
Python3
import itertools # initializing string test_string = 'GFG' # printing original string print ( "The original string : " + str (test_string)) # initializing K K = 'M' # No. of K required N = 5 # using repeat() # Append K character N times test_string + = "".join(itertools.repeat(K,N)) # print result print ( "The string after adding trailing K : " + str (test_string)) #This code is contributed by Edula Vinay Kumar Reddy |
The original string : GFG The string after adding trailing K : GFGMMMMM
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 5: Using a loop and concatenation
- Initialize a variable named test_string with the string value “GFG”.
- Initialize two variables, K with the string value “M” and N with the integer value 5.
- Start a for loop that will iterate N times, with the loop variable i taking on values from 0 to 4.
- Inside the loop, append the string value of K to test_string using the += operator. This will add K to the end of test_string each time the loop iterates.
- After the loop, print the string “The original string: ” concatenated with the string representation of test_string using the str() function.
- Print the string “The string after adding trailing K: ” concatenated with the string representation of test_string using the str() function. This will display the same value as the previous print statement since the test_string variable has not been modified since the loop.
Python3
test_string = "GFG" K = "M" N = 5 for i in range (N): test_string + = K print ( "The original string: " + str (test_string)) print ( "The string after adding trailing K: " + str (test_string)) |
The original string: GFGMMMMM The string after adding trailing K: GFGMMMMM
Time complexity: O(N) because the loop runs N times.
Auxiliary space: O(1) because only a few variables are used in addition to the input string.