Given a String, reverse shift each character according to its alphabetic position by K, including cyclic shift.
Input : test_str = ‘bccd’, K = 1
Output : abbc
Explanation : 1 alphabet before b is ‘a’ and so on.Input : test_str = ‘bccd’, K = 2
Output : zaab
Explanation : 2 alphabets before b is ‘z’ (rounded) and so on.
Method : Using maketrans() + upper() + list comprehension + translate() + slicing
In this, we make translation table to each character to its K shifted version using maketrans() and slicing. The upper() is used to handle all the upper case characters, translate() is used to perform translation according to lookup translation table created by maketrans().
Python3
# Python3 code to demonstrate working of # Reverse Shift characters by K# using maketrans() + upper() + list comprehension + translate() + slicing# initializing stringtest_str = 'GeeksForGeeks'# printing original Stringprint("The original string is : " + str(test_str))# initializing K K = 10alpha_chars = 'abcdefghijklmnopqrstuvwxyz'# converted to uppercasealpha_chars2 = alpha_chars.upper() # maketrans used for lowercase translationlower_trans = str.maketrans(alpha_chars, alpha_chars[ -K:] + alpha_chars[ : -K])# maketrans used for uppercase translationupper_trans = str.maketrans(alpha_chars2, alpha_chars2[ -K:] + alpha_chars2[ : -K])# merge lookupslower_trans.update(upper_trans)# make translation from lookupsres = test_str.translate(lower_trans)# printing result print("The converted String : " + str(res)) |
The original string is : GeeksForGeeks The converted String : WuuaiVehWuuai
The Time and Space Complexity for all the methods are the same:
Time Complexity: O(n)
Auxiliary Space: O(n)
Using ASCII values and modulo arithmetic in python:
Approach: We can use the ASCII values of the characters to shift them by K positions. We can subtract K from the ASCII value of each character and get the corresponding character. If the resulting ASCII value is less than the ASCII value of ‘a’, we can add 26 to it to get the correct character.
- Initialize an empty string result to store the shifted characters.
- Loop through each character char in the input string test_str.
- Convert the character char to its ASCII value using the ord function.
- Subtract the shift value K from the ASCII value. This will give the ASCII value of the shifted character.
- If the resulting ASCII value is less than the ASCII value of the character ‘a’, add 26 to it. This is because we want to wrap around to the end of the alphabet if we have shifted past the beginning of the alphabet.
- Convert the shifted ASCII value back to a character using the chr function.
- Append the shifted character to the result string.
- Return the result string containing the shifted characters.
Python3
# Python program for the above approach# Function to reverse the shift ASCII valuesdef reverse_shift_ascii(test_str, K): result = "" for char in test_str: ascii_val = ord(char) - K if ascii_val < ord('a'): ascii_val += 26 result += chr(ascii_val) return result# Driver Codetest_str_1 = 'bccd'K_1 = 1test_str_2 = 'bccd'K_2 = 2# Example outputsoutput_1 = 'abbc'output_2 = 'zaab'# Print the resultsprint("Input: test_str = '{}', K = {}".format(test_str_1, K_1))print("Output:", reverse_shift_ascii(test_str_1, K_1))print("Expected output:", output_1)print("Input: test_str = '{}', K = {}".format(test_str_2, K_2))print("Output:", reverse_shift_ascii(test_str_2, K_2))print("Expected output:", output_2) |
Input: test_str = 'bccd', K = 1 Output: abbc Expected output: abbc Input: test_str = 'bccd', K = 2 Output: zaab Expected output: zaab
Time Complexity: O(n), where n is the length of the string.
Space Complexity: O(n), since we are creating a new string of length n.
