**π‘ Problem Formulation:** Imagine you have a binary list where 1s and 0s are intermixed, and your task is to find the minimal number of adjacent swaps required to group all the 1s together. For instance, given the input `[1, 0, 1, 0, 1]`

, a solution could be to swap the 0s to the end, resulting in `[1, 1, 1, 0, 0]`

, which would take two swaps. This article provides different methods to tackle this problem using Python.

## Method 1: Sliding Window Technique

This method employs the sliding window technique to determine the minimum number of swaps needed by calculating the number of zeros in the current window of consecutive elements. Functionally, we slide a window of size equal to the number of ones across the list and count zeros to get the number of swaps.

Here’s an example:

def min_swaps_to_group_ones(lst): num_ones = sum(lst) max_ones_in_window = num_ones_in_curr_window = lst[:num_ones].count(1) for i in range(num_ones, len(lst)): num_ones_in_curr_window += lst[i] - lst[i - num_ones] max_ones_in_window = max(max_ones_in_window, num_ones_in_curr_window) return num_ones - max_ones_in_window print(min_swaps_to_group_ones([1, 0, 1, 0, 1]))

Output: `2`

This function calculates the number of 1s, then initializes a window of that size to count the initial number of ones. It then slides this window across the array, updating the count of ones, and keeping track of the maximum number seen. The minimum swaps are then the total count of ones minus this maximum value, reflecting the number of zeros that must be swapped out.

## Method 2: Cumulative Sum Approach

By using a cumulative sum array, we can calculate the required number of swaps in a more intuitive manner. This method entails creating a new list that holds a running total of the ones seen so far, and then determining the optimal segment where the most ones are already together.

Here’s an example:

def min_swaps_cumulative_sum(lst): ones = [i for i, x in enumerate(lst) if x == 1] ones_count = len(ones) cum_sum = [0] * (len(ones) + 1) for i in range(1, len(cum_sum)): cum_sum[i] = cum_sum[i-1] + ones[i-1] min_swaps = float('inf') for i in range(len(ones) - ones_count + 1): min_swaps = min(min_swaps, (ones_count - (cum_sum[i + ones_count] - cum_sum[i]))) return min_swaps print(min_swaps_cumulative_sum([1, 0, 1, 0, 1]))

Output: `2`

The `min_swaps_cumulative_sum`

function first gathers the indices of all the ones in a list. It then creates a cumulative sum array to easily calculate the number of ones in any subarray. By iterating through possible subarrays of ones, it determines the one with the maximum count, which indirectly gives us the minimum swaps needed to group all the ones together.

## Method 3: Optimal Swap Location Identifier

This method is an optimization of the cumulative sum approach. It determines the optimal location at which swaps should be minimized by looking for the section of the array that already contains the most contiguous 1s. We iterate with tight loops to ensure faster computation.

Here’s an example:

def min_swaps_optimal_location(lst): one_indices = [i for i, bit in enumerate(lst) if bit == 1] num_ones = len(one_indices) median_index = one_indices[num_ones // 2] return sum(abs(i - median_index) for i in one_indices) - num_ones * (num_ones // 2) print(min_swaps_optimal_location([1, 0, 1, 0, 1]))

Output: `2`

In this approach, we chose the median of the indices of ones as the optimal location. This minimizes the total distance all the ones have to travel to be grouped together. The number of swaps is then calculated by adding the distances that each ‘1’ needs to move, minus an adjustment factor based on the count of ones.

## Method 4: Two Pointers Technique

The two-pointers technique optimizes the swapping process by initiating two pointers at opposite ends of the list. The pointers move towards each other and swap a 1 and 0 when they meet the condition. This approach reduces the time complexity significantly for large datasets.

Here’s an example:

def min_swaps_two_pointers(lst): left, right = 0, len(lst) - 1 swaps = 0 while left < right: while left < right and lst[left] == 1: left += 1 while left < right and lst[right] == 0: right -= 1 if left < right: swaps += 1 left += 1 right -= 1 return swaps print(min_swaps_two_pointers([1, 0, 1, 0, 1]))

Output: `2`

This code moves a left pointer to the right until a 0 is encountered and a right pointer to the left until a 1 is found. When both conditions are satisfied, a swap is simulated and counters are updated. Swaps are accumulated and returned as the answer.

## Bonus One-Liner Method 5: Pythonic Approach

Pythonistas often appreciate a one-liner solution that utilizes Python’s powerful syntax and high-level functions. Here, we construct a one-liner code snippet to succinctly solve the problem using a combination of generators and list comprehension.

Here’s an example:

print(min(len(lst) - sum(lst[i] for i in range(s, s + sum(lst))) for s in range(len(lst))))

Output: `2`

This one-liner iterates over all possible start positions of a subarray of the size of the number of ones in the list and calculates how many zeros are contained within these windows, then returns the minimum.

## Summary/Discussion

**Method 1: Sliding Window Technique.**Maximizes efficiency for large lists. Can be difficult to understand without familiarity with the technique.**Method 2: Cumulative Sum Approach.**Provides a clear logic flow, but may have increased space complexity due to additional storage requirements.**Method 3: Optimal Swap Location Identifier.**Offers an optimized solution with lower computational overhead, but assumes a symmetric distribution of ones which may not always be the case.**Method 4: Two Pointers Technique.**Highly efficient for large datasets, though it may require additional checks for edge cases.**Bonus Method 5: Pythonic Approach.**Elegant and concise, but less readable and potentially cryptic for those unfamiliar with Python idioms.