5 Best Ways to Find the k-th Character of Decrypted String in Python

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πŸ’‘ Problem Formulation: The challenge involves decrypting a given encoded string and finding the k-th character of the resultant plaintext. For example, if the input string is “a2b3” which decrypts to “aabbb” and k = 4, the desired output is the fourth character of the decrypted string, which is “b”. The article provides different methods for achieving this in Python.

Method 1: Iterative Decryption

Iterative decryption involves processing each character of the encrypted string one by one, expanding them based on their encoded pattern, and storing the resulting decrypted string. Once the whole string is decrypted, the k-th character can be directly accessed. This method is straightforward, but it might not be efficient for very long strings where only a small part of the decrypted string is needed.

Here’s an example:

def iterative_decrypt(encoded_string, k):
    decrypted = ""
    i = 0
    while i < len(encoded_string):
        if encoded_string[i].isdigit():
            decrypted += decrypted[-1] * (int(encoded_string[i]) - 1)
        else:
            decrypted += encoded_string[i]
        i += 1
    return decrypted[k-1]

# Example usage:
print(iterative_decrypt('a2b3', 4))

Output: b

The function iterative_decrypt iterates through the encoded string, expanding each sequence as it goes and appending to the decrypted string. When the loop finishes, the function returns the k-th character from the decrypted string.

Method 2: Recursive Decryption

A recursive approach involves breaking down the problem into smaller sub-problems. It focuses on decrypting a part of the string and then recursively calling the function to decrypt the remaining part. Recursion can be more elegant but may be harder to understand and risks exceeding the maximum recursion depth for very large strings.

Here’s an example:

def recursive_decrypt(encoded_string, k):
    if k <= 0 or not encoded_string:
        return ''
    char, i, num_str = encoded_string[0], 1, ''
    while i < len(encoded_string) and encoded_string[i].isdigit():
        num_str += encoded_string[i]
        i += 1
    num = int(num_str) if num_str else 1
    length = num if char.isalpha() else 0
    if k <= length:
        return char
    return recursive_decrypt(encoded_string[i:], k-length)

# Example usage:
print(recursive_decrypt('a2b3', 4))

Output: b

This function, recursive_decrypt, decrypts the first character and its repetitions. If the k-th character is not within this decrypted part, it recursively decrypts the remainder with an updated value of k.

Method 3: Lazy Evaluation

Lazy evaluation defers the decryption of characters until they are needed. It computes the length that each character expands to and keeps a running sum to determine the position of the k-th character without fully decrypting the entire string. This method is memory-efficient and fast for large strings with large repetition factors.

Here’s an example:

def lazy_decrypt(encoded_string, k):
    count = 0
    for char in encoded_string:
        if char.isdigit():
            count *= int(char)
        else:
            count += 1
        if k <= count:
            return char if not char.isdigit() else prev_char
        prev_char = char

# Example usage:
print(lazy_decrypt('a2b3', 4))

Output: b

The lazy_decrypt function, uses a counter to track the expansion of characters and identifies the k-th character. No actual decryption of characters is performed unless they are needed to find the k-th character.

Method 4: Stack-Based Decryption

This method uses a stack data structure to simulate the decryption process. The idea is to push characters onto the stack and then multiply and re-insert them when a digit is found. The solution then simply counts the characters on the stack to find the k-th one. This approach is more intuitive for those familiar with stack operations.

Here’s an example:

def stack_decrypt(encoded_string, k):
    stack = []
    for char in encoded_string:
        if char.isalpha():
            stack.append(char)
        else:
            stack.extend([stack[-1]] * (int(char) - 1))
    return stack[k-1]

# Example usage:
print(stack_decrypt('a2b3', 4))

Output: b

The stack_decrypt function decodes the string by using a stack to handle characters and their frequencies. The function takes advantage of Python lists’ ability to be treated as stacks.

Bonus One-Liner Method 5: Using Regular Expressions

The one-liner approach uses the power of regular expressions to find and expand the digit-character pairs, essentially replicating the decryption in a functional programming style. It is a quick method but can become unreadable and may not be suitable for very complex encryption patterns.

Here’s an example:

import re
def oneliner_decrypt(encoded_string, k):
    return "".join(char * int(num) if num else char for char, num in re.findall(r'(\D)(\d*)', encoded_string))[k-1]

# Example usage:
print(oneliner_decrypt('a2b3', 4))

Output: b

This function, oneliner_decrypt, employs a list comprehension and regular expressions to match pairs of characters and their following digits, expanding them to form the decrypted string, and then returning the k-th character.

Summary/Discussion

  • Method 1: Iterative Decryption. Simple and explicit. Can be inefficient for decoding long strings when only one character is needed.
  • Method 2: Recursive Decryption. Elegant and concise. May result in stack overflow errors for very long strings.
  • Method 3: Lazy Evaluation. Memory-efficient and fast for large strings. May not be intuitive and requires understanding how lazy evaluation works.
  • Method 4: Stack-Based Decryption. Intuitive for those comfortable with stacks. Not the most efficient for long strings with many repeated characters.
  • Method 5: Using Regular Expressions. Quick and powerful for simple patterns. Can become complex and hard to maintain for more elaborate encoded strings.