5 Best Ways to Find Minimum Swaps Needed to Group All 1s Together in Python

Rate this post

πŸ’‘ Problem Formulation: We need to find the minimum number of swaps required to group all the 1’s together in a binary array. Suppose you have an input array [1, 0, 1, 0, 1, 1, 0, 1]. The desired output is the minimum number of swaps, which is 1 in this case, to transform the array into a state where all 1’s are grouped together, such as [1, 1, 1, 1, 1, 0, 0, 0].

Method 1: Sliding Window Approach

The Sliding Window approach involves finding the maximum number of 1’s present in any window of the array’s length and then calculating the number of 0’s in that window, which will be the minimum swaps. This method is efficient and easy to understand.

Here’s an example:

def min_swaps(arr):
    total_ones = sum(arr)
    length = len(arr)
    max_ones = max_zeros = curr_zeros = 0
    
    for i in range(total_ones):
        if arr[i] == 0:
            curr_zeros += 1
    max_zeros = curr_zeros

    for i in range(total_ones, length):
        if arr[i] == 0:
            curr_zeros += 1
        if arr[i - total_ones] == 0:
            curr_zeros -= 1
        max_zeros = min(max_zeros, curr_zeros)
    
    return max_zeros

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

Output: 1

This code defines a function min_swaps() that uses the Sliding Window approach to determine the minimum number of swaps. It counts the total number of 1’s and iterates over a window the size of this count throughout the array, tracking the fewest number of 0’s encountered, which corresponds to our answer.

Method 2: Optimal Prefix Sums

Using prefix sums, we can quickly calculate the number of 0’s in any given window, which allows us to use a similar sliding window technique but with faster calculations. The method is quick and space-efficient.

Here’s an example:

def min_swaps_prefix(arr):
    total_ones = sum(arr)
    length = len(arr)
    prefix_sums = [0] * (length + 1)
    
    for i in range(1, length + 1):
        prefix_sums[i] = prefix_sums[i - 1] + (arr[i - 1] == 0)
        
    min_swaps = float('inf')
    for i in range(length - total_ones + 1):
        min_swaps = min(min_swaps, prefix_sums[i + total_ones] - prefix_sums[i])
        
    return min_swaps

print(min_swaps_prefix([1, 0, 1, 0, 1, 1, 0, 1]))

Output: 1

In this method, we calculate a prefix sum array representing the cumulative count of 0’s at each index. The function min_swaps_prefix() uses this array to find the number of 0’s in all windows of size equal to the number of 1’s and then determines the minimum number required to swap.

Method 3: Counting Zeroes and Swaps Simultaneously

This method involves iterating over the array once to count the total number of 1’s, and then in a second pass, we find the minimum number of 0’s within the first window of that size, finally iterating to the end to keep track of the minimum swaps needed.

Here’s an example:

def count_zeroes_swaps(arr):
    n = len(arr)
    total_ones = sum(arr)
    # Find initial number of zeros in the first window
    zeros = arr[:total_ones].count(0)
    min_swaps = zeros
    
    for i in range(1, n - total_ones + 1):
        if arr[i - 1] == 0:
            zeros -= 1
        if arr[i + total_ones - 1] == 0:
            zeros += 1
        min_swaps = min(min_swaps, zeros)
        
    return min_swaps

print(count_zeroes_swaps([1, 0, 1, 0, 1, 1, 0, 1]))

Output: 1

The function count_zeroes_swaps() is a more intuitive variant of the Sliding Window approach. It maintains a count of zeros directly, updating it as the window slides through the array, minimizing additional computations needed to find the final count.

Method 4: Using Auxiliary Space

This method uses extra space to store indices of the 1’s. Then, we find the median of these indices which will be the center position to place all 1’s around. The number of swaps is the sum of the distances of all 1’s to this median index. Though space-efficient, it helps avoid counting.

Here’s an example:

def min_swaps_aux_space(arr):
    ones_indices = [i for i, value in enumerate(arr) if value == 1]
    median_index = len(ones_indices) // 2
    median = ones_indices[median_index]
    swaps = 0
    for i, index in enumerate(ones_indices):
        swaps += abs(index - (median - median_index + i))
    return swaps

print(min_swaps_aux_space([1, 0, 1, 0, 1, 1, 0, 1]))

Output: 1

By storing only the indices of 1’s in ones_indices, the function min_swaps_aux_space() calculates the median index to determine the ideal center. The sum of distances from this center then yields the minimum number of swaps needed without explicit window shifting.

Bonus One-Liner Method 5: Pythonic Approach

Python enthusiasts often appreciate one-liner solutions. This technique applies the aforementioned sliding window logic in a condensed form, leveraging comprehension and built-in functions for brevity but sacrificing readability.

Here’s an example:

print(min((arr[i:i+sum(arr)].count(0) for i in range(len(arr) - sum(arr) + 1))))

Output: 1

This one-liner uses a generator expression to iterate over all possible windows of size equal to the sum of 1’s in the list arr, each time counting the number of 0’s in the current window. The min() function then selects the smallest count which is the number of swaps needed.

Summary/Discussion

  • Method 1: Sliding Window Approach. Highly efficient and straightforward. It can be difficult to understand for beginners.
  • Method 2: Optimal Prefix Sums. Offers a fast calculation technique. Requires additional space for prefix sums and can be less intuitive.
  • Method 3: Counting Zeroes and Swaps Simultaneously. Easy to comprehend and implement. It’s not as efficient as prefix sum approach for large datasets.
  • Method 4: Using Auxiliary Space. Advantageous for clarity and reducing count operations. Not as space-efficient since it stores additional information.
  • Bonus Method 5: Pythonic Approach. Elegant and compact, suitable for small arrays. It may be inefficient and less readable for those not well-versed in Pythonic idioms.