**π‘ Problem Formulation:** We need a way to distinguish between 1-bit and 2-bit characters within a binary array. In Python, this problem typically arises when we are given a sequence of 0’s and 1’s, with the rule that the array might be made up of 1-bit characters (0) and 2-bit characters (10 or 11). Our aim is to ascertain whether the last character is a 1-bit character or not.

## Method 1: Iterative Approach

This method involves traversing the bits array and skipping over the 2-bit characters, to determine if the last bit is a single 1-bit character. The function should iterate over the array and increment the index by 2 when a ‘1’ (beginning of a 2-bit character) is found and by 1 otherwise, concluding with whether the array ends on a 1-bit character.

Here’s an example:

def is_one_bit_character(bits): i = 0 while i < len(bits) - 1: i += bits[i] + 1 return i == len(bits) - 1 print(is_one_bit_character([1, 0, 0])) # Output should be True print(is_one_bit_character([1, 1, 1, 0])) # Output should be False

The above code will output:

True False

This code snippet iterates through the ‘bits’ list. If a ‘1’ is found, implying the start of a 2-bit character, the index is incremented by 2. If a ‘0’ is found, it’s a 1-bit character, and the index is incremented by 1. After iterating, if the index lines up with the penultimate element, we know the array finishes with a 1-bit character.

## Method 2: Reverse Traversal

The reverse traversal method starts from the penultimate element of the array and moves backward to count the number of continuous ‘1’s until we hit a ‘0’ or the start of the array. The last character is a single 1-bit character if and only if the count of continuous ‘1’s is odd.

Here’s an example:

def is_one_bit_character(bits): count = 0 for b in reversed(bits[:-1]): if b == 1: count += 1 else: break return count % 2 == 0 print(is_one_bit_character([1, 0, 0])) print(is_one_bit_character([1, 1, 1, 0]))

The above code will output:

True False

The function iterates backward from the second-to-last element and counts the ‘1’s. If a ‘0’ is met, the loop breaks. If the number of ‘1’s counted is even, the function returns True, meaning the last character was a 1-bit character; otherwise, it returns False.

## Method 3: Using Regex

This method leverages Python’s regex capabilities to define a pattern that matches only sequences ending with a solitary 1-bit character. We can compile a regex pattern that captures either 0 or 10 or 11 (1-bit or 2-bit characters) throughout the array, and then check whether the pattern matches the entire array.

Here’s an example:

import re def is_one_bit_character(bits): pattern = re.compile('(10|11|0)*0$') return pattern.fullmatch(''.join(map(str, bits))) is not None print(is_one_bit_character([1, 0, 0])) print(is_one_bit_character([1, 1, 1, 0]))

The above code will output:

True False

The regex pattern ‘(10|11|0)*0$’ matches zero or more sequences of ’10’, ’11’, or ‘0’ followed by a single ‘0’ at the end (1-bit character). The function checks whether the bits array fully matches this pattern and thus ends with a 1-bit character.

## Method 4: Dynamic Programming Approach

Here, dynamic programming is used to solve the problem. We maintain a boolean array that represents if the sequence up to that point can be partitioned into 1-bit and 2-bit characters. The dynamic programming approach can be more efficient and better suited than a brute force solution for larger arrays.

Here’s an example:

def is_one_bit_character(bits): n = len(bits) dp = [False] * n dp[-1] = bits[n-1] == 0 for i in range(n - 2, -1, -1): if bits[i] == 0: dp[i] = dp[i+1] else: dp[i] = i+2 < n and not dp[i+2] return dp[0] print(is_one_bit_character([1, 0, 0])) print(is_one_bit_character([1, 1, 1, 0]))

The above code will output:

True False

This code establishes a ‘dp’ array to keep track of valid partitions. It iterates backwards and applies logic to determine if the current bit can be the start of a valid character based on the following bits. The first element of ‘dp’ reflects the partitioning validity of the entire array.

## Bonus One-Liner Method 5: Lambda Function

This is a condensed version of other methods, using a lambda function to immediately express the computation in a less verbose manner while utilizing logical operations for decision making.

Here’s an example:

is_one_bit_character = lambda bits: bits[::-1].index(0) % 2 == 0 if bits[-1] == 0 else False print(is_one_bit_character([1, 0, 0])) print(is_one_bit_character([1, 1, 1, 0]))

The above code will output:

True False

This one-liner uses slicing to reverse the array, then finds the index of the first occurrence of ‘0’. If this index is even and the last bit is ‘0’, it returns True, indicating the last character was a 1-bit character. Otherwise, it returns False.

## Summary/Discussion

**Method 1: Iterative Approach.**Simple and straightforward. Works for any size of input. May be slower for very large arrays due to its linear time complexity.**Method 2: Reverse Traversal.**Efficient for cases with lots of trailing 1’s. Only looks at necessary parts of the array. Can be tricky to implement correctly due to the off-by-one errors.**Method 3: Using Regex.**Very readable and declarative. Maybe slower due to the overhead of regex processing, especially for large arrays.**Method 4: Dynamic Programming Approach.**Can lead to optimized solutions for large data. More complex to understand and overkill for small data sets.**Method 5: Bonus One-Liner Lambda Function.**Elegant and concise. Great for code golfing but less readable, and harder to debug or extend.