**π‘ Problem Formulation:** The task is to determine if a binary string contains at most one continuous segment of ‘1’s or not. This means the program should return True if there are zero segments of ‘1’s or exactly one segment of ‘1’s, and False otherwise. For example, the input ‘110011’ has two separate segments of ‘1’s and should therefore return False while ‘11000’ with one segment should return True.

## Method 1: Iterative Check

This method involves iterating through the string and counting the number of segments of ones. If at the end of the iteration, the count is greater than one, return False. Otherwise, return True.

Here’s an example:

def has_at_most_one_segment_of_ones(binary_string): count = 0 i = 0 while i < len(binary_string): if binary_string[i] == '1': count += 1 while i < len(binary_string) and binary_string[i] == '1': i += 1 i += 1 return count <= 1 print(has_at_most_one_segment_of_ones('110011'))

Output:

False

The function `has_at_most_one_segment_of_ones()`

counts segments of ‘1’s by iterating over the binary string and checking for continuous sequences of ‘1’s. Once it encounters a ‘0’, it increases the segment counter and skips over the ones. If found more than one segment, the function returns False.

## Method 2: Using Regular Expressions

Regular expressions can be used for pattern matching. If the binary string matches the pattern of a zero or more occurrences of ‘0’, followed by a single segment of ‘1’s, followed by zero or more occurrences of ‘0’, it has at most one segment of ones.

Here’s an example:

import re def has_at_most_one_segment_of_ones_regex(binary_string): return bool(re.fullmatch('0*1*0*', binary_string)) print(has_at_most_one_segment_of_ones_regex('110011'))

Output:

False

The `has_at_most_one_segment_of_ones_regex()`

function uses the `re.fullmatch()`

to check if the entire binary string matches the pattern. The pattern allows any number of ‘0’s, followed by any number of ‘1’s, and ends with any number of ‘0’s.

## Method 3: Split and Check

In this method, split the binary string by ‘0’s and then check if one or fewer sub-strings of ‘1’s are present.

Here’s an example:

def has_at_most_one_segment_of_ones_split(binary_string): segments = binary_string.split('0') one_segments = [s for s in segments if '1' in s] return len(one_segments) <= 1 print(has_at_most_one_segment_of_ones_split('110011'))

Output:

False

The function `has_at_most_one_segment_of_ones_split()`

first splits the string into a list where each element is a possible segment of ones. It then filters out the segments that contain ‘1’s and counts them.

## Method 4: Find and RFind

This method leverages the `find()`

and `rfind()`

string methods to check the first and last occurrence of ‘1’. If both occurrences are within the same segment, the string satisfies the condition.

Here’s an example:

def has_at_most_one_segment_of_ones_find(binary_string): first_one = binary_string.find('1') last_one = binary_string.rfind('1') return first_one == -1 or last_one == first_one or '0' not in binary_string[first_one:last_one] print(has_at_most_one_segment_of_ones_find('110011'))

Output:

False

The function `has_at_most_one_segment_of_ones_find()`

checks if either there are no ones in the string (first_one == -1) or if all ones are within a single segment without ‘0’s between the first and last ‘1’.

## Bonus One-Liner Method 5: Exploiting Count

A Pythonic one-liner takes advantage of string methods. This one-line function checks the segment of ones by counting the occurrences of ’01’ or ’10’ as a segment delimiter.

Here’s an example:

has_at_most_one_segment_of_ones_oneliner = lambda s: s.count('01') + s.count('10') <= 1 print(has_at_most_one_segment_of_ones_oneliner('110011'))

Output:

False

This lambda function checks if there is at most one segment of ones by counting transition patterns that would indicate the start or end of a ‘1’s segment.

## Summary/Discussion

**Method 1: Iterative Check.**Easy to understand. Short-circuits on finding a second segment. Inefficient for long strings without ‘0’.**Method 2: Using Regular Expressions.**Compact and powerful. Readability may suffer for those unfamilar with regex. Pattern matching can be slow on very large strings.**Method 3: Split and Check.**Very readable. Utilizes built-in string methods efficiently. Additional memory usage for the list of segments.**Method 4: Find and RFind.**Straightforward concept relying on string find functions. Efficient if ‘1’s are clustered near the start or end. Can be inefficient if ‘1’s are in the middle.**Bonus One-Liner Method 5:**Extremely concise. Relies on two counts, which is less efficient. Clear and clever use of string methods.