5 Best Ways to Find Minimum Absolute Sum Difference in Python

💡 Problem Formulation: The task is to find the minimum absolute sum difference between two arrays of equal length. Given arrays arr1 and arr2, the goal is to compute the sum of absolute differences for each pair of elements, and then return the smallest possible sum after modifying at most one element from arr1. For example, input arrays [1,3,5] and [2,4,6] would yield an output of 3.

Method 1: Brute Force Approach

This method involves iterating through each element in the first array, substituting it with each possible element from the second array and computing the sum of absolute differences every time to find the minimum. While straightforward, this approach is not efficient for large arrays due to its O(n^2) complexity.

Here’s an example:

def min_absolute_sum_diff(arr1, arr2):
    result = sum(abs(a - b) for a, b in zip(arr1, arr2))
    for i in range(len(arr1)):
        for j in range(len(arr2)):
            new_diff = abs(arr1[i] - arr2[j])
            result = min(result, sum(abs(a - b) if k != i else new_diff for k, (a, b) in enumerate(zip(arr1, arr2))))
    return result

arr1 = [1,3,5]
arr2 = [2,4,6]
print(min_absolute_sum_diff(arr1, arr2))

The output of this code snippet:

3

The code defines a function min_absolute_sum_diff that will check every possible substitute from arr1 with elements from arr2 and calculate the new sum. Finally, it returns the smallest sum it found throughout the loop.

Method 2: Sorting with Binary Search

This method improves performance by first sorting the second array and using binary search to find the closest element to each element in the first array, thereby minimizing the sum at each step. This method’s time complexity is O(n log n) due to the sorting step.

Here’s an example:

import bisect

def min_absolute_sum_diff(arr1, arr2):
    arr2_sorted = sorted(arr2)
    result = current = sum(abs(a - b) for a, b in zip(arr1, arr2))
    for a, b in zip(arr1, arr2):
        index = bisect.bisect_left(arr2_sorted, a)
        if index  0:
            result = min(result, current - abs(a - b) + abs(arr2_sorted[index - 1] - b))
    return result

arr1 = [1,3,5]
arr2 = [2,4,6]
print(min_absolute_sum_diff(arr1, arr2))

The output of this code snippet:

3

In this example, bisect.bisect_left is used to find the position to insert each element from arr1 into arr2_sorted, to find the closest values and minimize the absolute sum difference efficiently.

Method 3: Use of a Heuristic

This method includes employing a heuristic to minimize the recalculations required for elements that are close to each other within the array. It can often reduce complexity in practice but does not guarantee an optimal solution in all cases. It’s more of an optimization technique rather than a direct algorithm.

Here’s an example:

def min_absolute_sum_diff(arr1, arr2):
    # Placeholder for the heuristic method
    pass
    
# Since this is a heuristic method, a proper example is very contextual and would be developed based on the specific heuristic chosen.

The output would be variable depending on the heuristic applied.

This placeholder function demonstrates where one would implement a heuristic method to find the minimum absolute sum difference. The implementation of this is highly context-dependent, hence it’s only illustrative.

Method 4: Dynamic Programming

This method uses dynamic programming to build a solution incrementally, making use of previous calculations to find the minimum absolute sum difference. This can be very efficient for certain types of input data but can also be overkill for simpler problems or where data is not amenable to dynamic optimization.

Here’s an example:

def min_absolute_sum_diff(arr1, arr2):
    # Placeholder for the dynamic programming method
    pass
    
# Since this is a dynamic programming method, the example would depend heavily on the recurrence relation developed for the specific problem instance.

The output would be dependent on the approach taken for dynamic programming.

Similarly, this function represents a template for a dynamic programming approach. The actual implementation would vary significantly based on the specific properties of the input data and the recurrence relation designed.

Bonus One-Liner Method 5: Practical Compromise

For scenarios where readability and simplicity take precedence over performance, a concise one-liner may be employed using Python’s built-in functions and list comprehensions.

Here’s an example:

min_absolute_sum_diff = lambda arr1, arr2: min(sum(abs(a - b) for a, b in zip(arr1, arr2)), sum(abs(a - min(arr2, key=lambda x: abs(x - a))) for a in arr1))
arr1 = [1,3,5]
arr2 = [2,4,6]
print(min_absolute_sum_diff(arr1, arr2))

The output of this code snippet:

3

This one-liner lambda function attempts to minimize the sum of absolute differences by either keeping the original pairs or pairing each element from arr1 with the closest element in arr2, evaluated inline using min and a key function.

Summary/Discussion

• Method 1: Brute Force Approach. It’s intuitive and straightforward but the least efficient, best suited for small datasets due to its O(n^2) complexity.
• Method 2: Sorting with Binary Search. Offers a significant performance boost, making it the best general-purpose solution with O(n log n) complexity.
• Method 3: Use of a Heuristic. It can provide faster solutions on large datasets with certain patterns, but lacks a one-size-fits-all formula and doesn’t ensure optimality.
• Method 4: Dynamic Programming. Optimizes the problem-solving process for specific types of data but may introduce unnecessary complexity for simpler problems.
• Method 5: Practical Compromise One-Liner. Prioritizes code simplicity and readability over performance, suitable for quick and dirty solutions where execution speed is not critical.