Python – Reverse Shift characters by K

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 string
test_str = 'GeeksForGeeks'

# printing original String
print("The original string is : " + str(test_str))

# initializing K
K = 10

alpha_chars = 'abcdefghijklmnopqrstuvwxyz'

# converted to uppercase
alpha_chars2 = alpha_chars.upper()

# maketrans used for lowercase translation
lower_trans = str.maketrans(alpha_chars, alpha_chars[ -K:] + alpha_chars[ : -K])

# maketrans used for uppercase translation
upper_trans = str.maketrans(alpha_chars2, alpha_chars2[ -K:] + alpha_chars2[ : -K])

# merge lookups
lower_trans.update(upper_trans)

# make translation from lookups
res = test_str.translate(lower_trans)

# printing result
print("The converted String : " + str(res))

Output

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.

1. Initialize an empty string result to store the shifted characters.
2. Loop through each character char in the input string test_str.
3. Convert the character char to its ASCII value using the ord function.
4. Subtract the shift value K from the ASCII value. This will give the ASCII value of the shifted character.
5. 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.
6. Convert the shifted ASCII value back to a character using the chr function.
7. Append the shifted character to the result string.
8. Return the result string containing the shifted characters.

Python3

# Python program for the above approach

# Function to reverse the shift ASCII values
def 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 Code
test_str_1 = 'bccd'
K_1 = 1

test_str_2 = 'bccd'
K_2 = 2

# Example outputs
output_1 = 'abbc'
output_2 = 'zaab'

# Print the results
print("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)

Output

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.