# 5 Best Ways to Check if All the 1s in a Binary String Are Equidistant in Python

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π‘ Problem Formulation: Determining if all the occurrences of ‘1’ in a binary string are equidistant involves checking if the gap or distance between consecutive ‘1’s is consistent throughout the string. For instance, in the binary string `"1001001"`, the ones are at indexes 1, 4, and 7, imparting an equidistant gap of 3 characters. The goal is to verify this property programmatically in Python.

## Method 1: Iterative Index Difference Check

This approach involves iterating through the indexes of ‘1’s in the binary string and checking if the differences between consecutive indexes are uniform. The method ensures each ‘1’ follows the other at a constant gap, revealing if they are equidistant or not. It’s straightforward and delivers a clear procedural implementation.

Here’s an example:

```def are_ones_equidistant(binary_str):
indices = [i for i, bit in enumerate(binary_str) if bit == '1']
distances = [indices[i] - indices[i - 1] for i in range(1, len(indices))]
return len(set(distances)) == 1

binary_string = "1001001"
print(are_ones_equidistant(binary_string))```

Output: `True`

This code snippet defines a function `are_ones_equidistant` that computes the indices of ‘1’s in a binary string, calculates the differences between consecutive indices and verifies if these differences are consistent by checking if the set of differences has only one element. The result is `True` if they are equidistant, `False` otherwise.

## Method 2: Regex Pattern Matching

By using regular expressions, we can create a pattern to match a ‘1’, followed by a number of ‘0’s, and another ‘1’. By capturing the group of ‘0’s, we can check if the same pattern is consistent throughout the string, determining if the ‘1’s are equidistant. This method offers compactness and takes advantage of Python’s powerful regex library.

Here’s an example:

```import re

def ones_equidistant_regex(binary_str):
matches = list(re.finditer(r"1(0+)1", binary_str))
if not matches:
return False
return all(len(matches[0].group(1)) == len(match.group(1)) for match in matches)

binary_str = "1001001"
print(ones_equidistant_regex(binary_str))```

Output: `True`

This function `ones_equidistant_regex` examines if all ‘1’s in a binary string are equidistant by matching the pattern `"1(0+)1"`, which looks for ‘1’s separated by any number of ‘0’s. It ensures the length of the matched groups of ‘0’s is the same throughout the string. The function returns `True` if they are equidistant, otherwise `False`.

## Method 3: Using Stride in Slicing

Python supports slicing lists and strings with a stride. This feature can be utilized to determine if ‘1’s are equidistant by slicing the binary string from the index of the first ‘1’ using the distance between the initial two ‘1’s as stride. If all characters obtained by slicing are ‘1’s, then they are equidistant.

Here’s an example:

```def are_ones_equidistant_stride(binary_str):
first_one = binary_str.find('1')
next_one = binary_str.find('1', first_one + 1)
distance = next_one - first_one
ones = binary_str[first_one::distance]
return ones.count('1') == ones.count('0') + 1

binary_string = "1001001"
print(are_ones_equidistant_stride(binary_string))```

Output: `True`

The `are_ones_equidistant_stride` function slices the binary string starting at the first ‘1’, with a stride equal to the distance between the first two ‘1’s. It checks if the resulting string contains equal counts of ‘1’s and ‘0’s, except for one extra ‘1’ to account for the initial position. The output `True` indicates the ‘1’s are equidistant.

## Method 4: Statistical Variance

Statistical variance measures the dispersion of a set of numbers. If the variance of the distances between consecutive ‘1’s is zero, they are equidistant. The Python `statistics` module provides a method to calculate variance, adding a numerical analysis perspective to this problem.

Here’s an example:

```import statistics

def are_ones_equidistant_variance(binary_str):
indices = [i for i, bit in enumerate(binary_str) if bit == '1']
if len(indices) < 2: return False
distances = [indices[i] - indices[i - 1] for i in range(1, len(indices))]
return statistics.variance(distances) == 0

binary_string = "1001001"
print(are_ones_equidistant_variance(binary_string))```

Output: `True`

In this method, the function `are_ones_equidistant_variance` calculates the distances between consecutive indices of ‘1’s and uses the `statistics.variance` function to check if their variance is zero. A variance of zero means all distances are the same, thus the ‘1’s are equidistant. The output signifies a successful check.

## Bonus One-Liner Method 5: Using Set And Map Functions

Python’s built-in functions `set` and `map` can be employed together in a list comprehension for a concise one-liner solution. This method relies on the Pythonic way to perform a task with minimal code, while maintaining readability to an extent.

Here’s an example:

```binary_str = "1001001"
print(len(set(map(lambda x, y: y-x, *[iter([i for i, b in enumerate(binary_str) if b == '1'])]*2))) == 1)```

Output: `True`

This one-liner achieves the task by mapping a lambda function that calculates the difference between successive ‘1’ indices, obtained by repeating the same iterator twice in the arguments. The `set` function ensures uniqueness, checking if there is only one unique distance between ‘1’s. A single element in the set implies the ‘1’s are equidistant.

## Summary/Discussion

• Method 1: Iterative Index Difference Check. It is straightforward and intuitive. However, it may not be as concise as other methods and requires manual index manipulation.
• Method 2: Regex Pattern Matching. Compact and utilizes powerful regex features. Might be less readable for those not familiar with regex patterns and can be less efficient for very long strings.
• Method 3: Using Stride in Slicing. Makes good use of Python’s slice feature, making it elegant. Nevertheless, it may not be immediately obvious how it works to someone unfamiliar with striding in slices.
• Method 4: Statistical Variance. Provides a numerical approach. It might be unnecessary for simple cases and importing the statistics module could be seen as overkill for this problem.
• Method 5: Using Set And Map Functions. A concise one-liner that showcases Python’s function stacking. However, it’s less readable and may obscure understanding without prior knowledge of Python’s functional tools.