Java Program To Find Sum Of Minimum Absolute Difference Of The Given Array

Finding the sum of minimum absolute differences within an array is a common problem in computer science that often involves sorting to achieve an optimal solution. It tests understanding of array manipulation and the properties of absolute values.

In this article, you will learn how to efficiently calculate the sum of minimum absolute differences in a given integer array using Java.

Problem Statement

The problem is to find the sum of absolute differences between adjacent elements after arranging the array elements to minimize this sum. Specifically, given an array of integers, we need to reorder its elements such that the sum of |arr[i] - arr[i+1]| for all i from 0 to n-2 is minimized, and then calculate this minimum sum.

This problem often arises in scenarios requiring optimal arrangement, such as scheduling tasks, minimizing material waste in a production line, or optimizing data transfer order to reduce variance between consecutive packets. The goal is to reduce overall 'jumps' or discrepancies between successive items.

Example

Consider the array [5, 2, 8, 3]. If we don't sort, one possible arrangement is [5, 2, 8, 3]: |2-5| + |8-2| + |3-8| = 3 + 6 + 5 = 14

If we sort the array, it becomes [2, 3, 5, 8]: |3-2| + |5-3| + |8-5| = 1 + 2 + 3 = 6

The minimum sum of absolute differences is 6.

Background & Knowledge Prerequisites

To understand and implement the solution, you should be familiar with:

• Java Basics: Variables, loops, arrays, and methods.
• Absolute Value: Understanding Math.abs() function.
• Sorting Algorithms: Basic knowledge of how sorting works (e.g., Arrays.sort() in Java).

You will need the java.util.Arrays class for its sorting utility.

Use Cases or Case Studies

This problem has several practical applications:

• Manufacturing Optimization: In assembly lines, parts might need to be processed in an order that minimizes changes in machine settings between consecutive parts. Sorting the parts based on a critical dimension can reduce total adjustment time.
• Data Compression: Algorithms might try to reorder data elements to reduce the differences between adjacent values, which can improve the efficiency of certain differential encoding schemes.
• Financial Data Analysis: Analyzing stock price volatility. While not a direct application of this exact problem, understanding how to minimize differences between sequential data points is a fundamental concept in time series analysis to identify trends or reduce noise.
• Route Planning: While complex, a simplified version of minimizing differences between adjacent stops (e.g., by some metric like altitude or fuel requirement) could be a component of an optimized route.
• Image Processing: In image filtering, processing pixels in an order that minimizes color differences between adjacent pixels could be relevant for certain noise reduction or smoothing algorithms.

Solution Approaches

For the problem of finding the sum of minimum absolute differences of an array, there is one primary and optimal approach: sorting the array first.

Sorting and Summing Adjacent Differences

One-line summary: Sort the array and then iterate through the sorted array, summing the absolute differences of all adjacent elements.

The key insight is that to minimize the sum of absolute differences between adjacent elements in a sequence, the array must be sorted. If the array is unsorted, there will always be at least two elements a[i] and a[j] such that a[i] < a[k] < a[j] for some k between i and j (or vice-versa). Swapping elements to bring a[k] adjacent to either a[i] or a[j] (or both by sorting the whole segment) will reduce the overall sum of differences. When the array is sorted, any element a[i] is closest in value to its immediate neighbors a[i-1] and a[i+1].

// Sum of Minimum Absolute Differences
import java.util.Arrays; // Required for sorting the array

// Main class containing the entry point of the program
public class Main {
    public static void main(String[] args) {
        // Step 1: Define the input array
        int[] arr = {5, 2, 8, 3};
        System.out.println("Original Array: " + Arrays.toString(arr));

        // Step 2: Sort the array
        // Sorting ensures that elements closest in value are adjacent,
        // which minimizes their absolute difference.
        Arrays.sort(arr);
        System.out.println("Sorted Array:   " + Arrays.toString(arr));

        // Step 3: Initialize a variable to store the sum of absolute differences
        long minAbsoluteDiffSum = 0; // Use long to handle potentially large sums

        // Step 4: Iterate through the sorted array and calculate the sum of
        //         absolute differences between adjacent elements
        for (int i = 0; i < arr.length - 1; i++) {
            // Calculate the absolute difference between the current element and the next
            long diff = Math.abs((long)arr[i] - arr[i + 1]);
            minAbsoluteDiffSum += diff;
        }

        // Step 5: Print the calculated sum
        System.out.println("Sum of minimum absolute differences: " + minAbsoluteDiffSum);

        // Example with another array
        int[] arr2 = {1, 10, 5, 2, 8};
        System.out.println("\\nOriginal Array 2: " + Arrays.toString(arr2));
        Arrays.sort(arr2);
        System.out.println("Sorted Array 2:   " + Arrays.toString(arr2));
        long minAbsoluteDiffSum2 = 0;
        for (int i = 0; i < arr2.length - 1; i++) {
            minAbsoluteDiffSum2 += Math.abs((long)arr2[i] - arr2[i + 1]);
        }
        System.out.println("Sum of minimum absolute differences for arr2: " + minAbsoluteDiffSum2);
    }
}

Run

Sample Output:

Original Array: [5, 2, 8, 3]
Sorted Array:   [2, 3, 5, 8]
Sum of minimum absolute differences: 6

Original Array 2: [1, 10, 5, 2, 8]
Sorted Array 2:   [1, 2, 5, 8, 10]
Sum of minimum absolute differences for arr2: 9

Stepwise Explanation:

1. Define Array: An integer array arr is initialized with the given numbers.
2. Sort Array: The Arrays.sort(arr) method is called. This sorts the array in ascending order. This step is crucial because it places elements with the smallest differences next to each other.
3. Initialize Sum: A long variable minAbsoluteDiffSum is initialized to 0. long is used to prevent potential integer overflow if the sum of differences becomes very large.
4. Iterate and Sum: A for loop iterates from the first element up to the second-to-last element (arr.length - 1). In each iteration:

• Math.abs((long)arr[i] - arr[i + 1]) calculates the absolute difference between the current element arr[i] and its adjacent element arr[i + 1]. Casting to long before subtraction helps prevent overflow during the subtraction itself if arr[i] and arr[i+1] are large integers.
• This difference is added to minAbsoluteDiffSum.

1. Print Result: After the loop completes, minAbsoluteDiffSum holds the total sum of minimum absolute differences, which is then printed to the console.

Time and Space Complexity:

• Time Complexity: The dominant operation is sorting the array, which typically takes O(N log N) time, where N is the number of elements in the array. The subsequent loop to sum differences takes O(N) time. Therefore, the overall time complexity is O(N log N).
• Space Complexity: Arrays.sort() in Java uses a dual-pivot quicksort for primitive types, which has an average space complexity of O(log N) for the recursion stack (though it can be O(N) in the worst case for object arrays). For primitive arrays, it's generally considered in-place or O(log N). If a custom sort that uses O(N) space is used, then the space complexity would be O(N). For this specific problem, it's typically O(log N) or O(1) (in-place) depending on the specific Arrays.sort implementation details for primitives.

Conclusion

Calculating the sum of minimum absolute differences in an array is an elegant problem solved most efficiently by first sorting the array. By arranging elements in ascending order, we guarantee that adjacent elements will have the smallest possible absolute difference, thus minimizing their cumulative sum. This approach is simple to implement and offers an optimal time complexity of O(N log N) due to the sorting step.

Summary

• The problem aims to minimize the sum of absolute differences between adjacent elements.
• The optimal strategy involves sorting the array first.
• Once sorted, iterate through the array, summing the Math.abs(arr[i] - arr[i+1]) for all adjacent pairs.
• This method guarantees the minimum possible sum because sorting places elements with the smallest numerical gap next to each other.
• The solution has a time complexity of O(N log N) primarily due to the sorting step.