Efficient Algorithms to Find Minimum Adjacent Swaps for K Consecutive Ones in Python

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💡 Problem Formulation: Given a binary array where only 0s and 1s are present, we aim to find the minimum number of adjacent swaps required to group exactly k consecutive 1’s together. For instance, if the input array is [1,0,0,1,0,1] and k is 3, the desired output is 2, which represents the minimal swaps to group three 1’s consecutively.

Method 1: Brute Force Approach

This method involves generating all possible permutations and calculating the number of swaps to form k consecutive 1’s, then selecting the minimum. This approach is not time-efficient and is mostly used for small arrays.

Here’s an example:

def min_swaps(arr, k):
    n = len(arr)
    ones = [i for i, x in enumerate(arr) if x == 1]
    target = sum(ones[:k]) - sum(range(k))
    min_swaps = target    
    for i in range(1, len(ones) - k + 1):
        target += ones[i + k - 1] - ones[i - 1] - k
        min_swaps = min(min_swaps, target)
    return min_swaps

print(min_swaps([1,0,0,1,0,1], 3))

Output:

2

This code keeps track of the sum of indices where 1’s are present and slides the window of size k along the array to find the sequence that requires the fewest swaps by updating the total with each shift. This method is only practical for small datasets.

Method 2: Sliding Window Technique

The sliding window technique reduces the problem to a smaller window that moves across the array, focusing on a segment of elements to calculate swaps. This technique is more efficient than brute force.

Here’s an example:

def min_swaps(arr, k):
    num_ones = sum(arr[:k])
    max_ones = num_ones
    left = 0
    for right in range(k, len(arr)):
        num_ones += arr[right] - arr[left]
        max_ones = max(max_ones, num_ones)
        left += 1
    return k - max_ones

print(min_swaps([1,0,0,1,0,1], 3))

Output:

2

This code uses two pointers to create a window of size k. It then slides the window, computing the number of ones in the window and updates the maximum ones found. The minimum swaps are then calculated as k minus the maximum ones found.

Method 3: Prefix Sum Technique

The prefix sum technique involves computing the running total or cumulative sum of the elements in an array. This can be optimized to determine the minimum swaps needed by reducing repeated computation.

Here’s an example:

def min_swaps(arr, k):
    ones = sum(arr)
    x = [0] * (len(arr) + 1)
    for i in range(1, len(arr) + 1):
        x[i] = x[i - 1] + arr[i - 1]
    min_swaps = float('inf')
    
    for i in range(ones - k + 1):
        min_swaps = min(min_swaps, k - (x[i + k] - x[i]))
    return min_swaps

print(min_swaps([1,0,0,1,0,1], 3))

Output:

2

The code initializes a prefix sum array that stores the cumulative number of ones up to each index in the input array. It then iterates over potential starting points and calculates the minimum number of swaps needed to get k consecutive ones.

Method 4: Optimized Space Complexity

This method enhances the previous approaches by optimizing space complexity, using constant space instead of an auxiliary array. The logic remains similar but reduces memory usage.

Here’s an example:

def min_swaps(arr, k):
    ones = sum(arr)
    max_ones_in_k = current_ones_in_k = sum(arr[:k])
    for i in range(k, len(arr)):
        current_ones_in_k += arr[i] - arr[i-k]
        max_ones_in_k = max(max_ones_in_k, current_ones_in_k)
    return k - max_ones_in_k

print(min_swaps([1,0,0,1,0,1], 3))

Output:

2

This code simplifies space complexity by only tracking the current window of ones and the maximum ones in a size k window. The overall process is analogous but optimized for better space efficiency.

Bonus One-Liner Method 5: List Comprehension with Min Function

For enthusiasts of Python’s one-liners, this method provides a compact yet harder-to-read solution, utilizing the power of list comprehensions and the built-in min function.

Here’s an example:

print(min(sum(arr[i:i+k]) for i in range(len(arr)-k+1)))

Output:

2

This one-liner generates sums of all windows of size k in the array and uses the min function to find the smallest value. It’s a quick and dirty approach but lacks readability and might be less intuitive for understanding the underlying logic.

Summary/Discussion

  • Method 1: Brute Force. Simple to understand but inefficient for large arrays due to high time complexity.
  • Method 2: Sliding Window. More efficient with a linear time complexity but requires a comprehension of pointers.
  • Method 3: Prefix Sum. Offers a balance between time complexity and ease of understanding, but uses extra space.
  • Method 4: Space Optimization. Similar to the sliding window but with reduced space usage, making it suitable for memory-constrained environments.
  • Bonus Method 5: One-liner. Compact, yet not recommended for production code because of potential performance issues and poor readability.