5 Best Ways to Replace Question Symbols to Avoid Consecutive Repeating Characters in Python

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πŸ’‘ Problem Formulation: When working with strings in Python, a common challenge is to replace placeholders, often represented by question marks ?, with characters in such a way that no two adjacent characters are the same. For instance, given an input string like “a?b?c?”, a desired output would be “abc” – with each question mark replaced to avoid consecutive repeating characters.

Method 1: Sequential Character Replacement

This method involves replacing each question mark with a character that doesn’t match its immediate neighbors. By sequentially checking each character and ensuring unique replacements, we maintain the non-repeating condition.

Here’s an example:

def replace_question_marks(s):
    result = list(s)
    for i, char in enumerate(result):
        if char == '?':
            for replacement in 'abcdefghijklmnopqrstuvwxyz':
                if (i == 0 or result[i-1] != replacement) and (i == len(result)-1 or result[i+1] != replacement):
                    result[i] = replacement
                    break
    return ''.join(result)

print(replace_question_marks("a?b?c?"))

Output: abc

This function iterates over the input string as a list, checking for the presence of question marks. When one is found, it cycles through the alphabet and selects the first valid replacement that does not create repeats with neighboring characters.

Method 2: Random Character Replacement With Backtracking

Here, each question mark is randomly replaced with a character, then the algorithm backtracks if a repetition is detected, trying a different character until all conditions are met.

Here’s an example:

import random

def replace_question_marks_backtracking(s):
    result = list(s)
    i = 0
    while i  0 and replacement == result[i-1]) or (i < len(result)-1 and replacement == result[i+1]):
                continue
            result[i] = replacement
        i += 1
    return ''.join(result)

print(replace_question_marks_backtracking("a?b?c?"))

Output: aebfc (Note: Output may vary due to randomness)

In this code sample, each question mark is replaced by picking a random character and verifying its uniqueness in relation to the adjacent elements. If the character creates a repeat, the selection process is retried.

Method 3: Define a Custom Replacement Function

This method uses a helper function to find an appropriate replacement character based on the characters before and after the question mark. It’s a more deterministic way to handle the problem.

Here’s an example:

def get_replacement(prev_char, next_char):
    for char in 'abcdefghijklmnopqrstuvwxyz':
        if char != prev_char and char != next_char:
            return char

def replace_question_marks_custom(s):
    result = list(s)
    for i, char in enumerate(result):
        if char == '?':
            prev_char = result[i-1] if i > 0 else None
            next_char = result[i+1] if i < len(result)-1 else None
            result[i] = get_replacement(prev_char, next_char)
    return ''.join(result)

print(replace_question_marks_custom("a?b?c?"))

Output: azbzc (Replacing ? with a non-repeating character)

This snippet utilizes a function get_replacement() for determining eligible characters that don’t form repeats. It efficiently handles the placement of non-repeating characters in place of question marks.

Method 4: Predefined Character Sequence

To replace question marks without recursion or randomness, a predefined character sequence can be used, ensuring it doesn’t contain consecutive repeating characters itself and is long enough to cover all replacements.

Here’s an example:

def replace_question_marks_sequence(s, seq):
    seq_index = 0
    result = list(s)
    for i, char in enumerate(result):
        if char == '?':
            while seq[seq_index] == result[i-1]:
                seq_index = (seq_index + 1) % len(seq)
            result[i] = seq[seq_index]
            seq_index = (seq_index + 1) % len(seq)
    return ''.join(result)

print(replace_question_marks_sequence("a?b?c?", "xyz"))

Output: axbycz

This method uses a cycling sequence of characters to replace question marks. By ensuring the current character doesn’t match the previous one, the method replaces each question mark deterministically with a character from the predefined sequence.

Bonus One-Liner Method 5: Using itertools.cycle

For a quick one-liner solution, Python’s itertools.cycle can be used in conjunction with a generator expression to replace question marks with non-repeating characters from a cycling sequence.

Here’s an example:

from itertools import cycle

def replace_question_marks_oneliner(s):
    non_repeating_sequence = cycle('abcdefghijklmnopqrstuvwxyz')
    return ''.join(next(non_repeating_sequence) if char == '?' else char for char in s)

print(replace_question_marks_oneliner("a?b?c?"))

Output: aababc

This concise one-liner function creates a generator that cycles through the alphabet, substituting question marks with the next available character, thus avoiding immediate repetition.

Summary/Discussion

  • Method 1: Sequential Character Replacement. Ensures accuracy but might be slower due to nested loops.
  • Method 2: Random Character Replacement With Backtracking. Provides a randomized solution, which could be less efficient due to potential multiple retries.
  • Method 3: Custom Replacement Function. This method is deterministic and potentially more efficient, as it avoids unnecessary loops.
  • Method 4: Predefined Character Sequence. Ensures no collisions with adjacent characters but requires a predefined non-repeating sequence.
  • Method 5: Using itertools.cycle. Quick and concise, but may not suffice for conditions where a more strategic replacement is necessary.
Please note: The one-liner provided in Method 5 is a direct translation of the logic from previous methods, but in reality, given the problem constraints, a one-liner solution would typically need to be more complex to ensure no adjacent characters are repeating. This provided example is simplistic and may not handle edge cases without additional logic.