Introduction:

The Optimal Merge Pattern is a classic problem that involves merging multiple sorted files into a single sorted file in the most efficient manner. The goal is to minimize the total computation time required for merging.

Conceptual Explanation:

Imagine you have several sorted lists, and you need to combine them into one sorted list. The challenge is to perform this merge operation with the least cost, where the cost is typically defined as the number of comparisons or the time taken to merge the lists.

Real-world Analogy:

Consider a librarian who wants to merge several sorted piles of books into a single sorted shelf. The librarian wants to minimize the effort (or time) spent in comparing and placing books in the correct order.

Algorithm Overview:

The optimal way to merge involves using a priority queue (min-heap) to always merge the two smallest lists first. This approach ensures that the overall cost is minimized.

Scenario:

Suppose we have files of sizes: 20, 30, 10, and 5 units.

Solution Approach:

Using a min-heap, we first merge the smallest files (5 and 10), resulting in a new file of size 15. Next, we merge 15 with 20, and finally, the result with 30.

Code Example:

import java.util.PriorityQueue;

public class OptimalMergePattern {
    public static void main(String[] args) {
        int[] files = {20, 30, 10, 5};
        PriorityQueue minHeap = new PriorityQueue<>();
        
        for (int file : files) {
            minHeap.add(file);
        }
        
        int totalCost = 0;
        while (minHeap.size() > 1) {
            int first = minHeap.poll();
            int second = minHeap.poll();
            int merged = first + second;
            totalCost += merged;
            minHeap.add(merged);
        }
        
        System.out.println("Total cost of merging: " + totalCost);
    }
}
            

Scenario:

Consider files of equal size: 10, 10, 10, and 10 units.

Solution Approach:

Even with equal sizes, the strategy remains the same: merge the smallest pairs first. This ensures the least cumulative merge cost.

Code Example:

import java.util.PriorityQueue;

public class OptimalMergePattern {
    public static void main(String[] args) {
        int[] files = {10, 10, 10, 10};
        PriorityQueue minHeap = new PriorityQueue<>();
        
        for (int file : files) {
            minHeap.add(file);
        }
        
        int totalCost = 0;
        while (minHeap.size() > 1) {
            int first = minHeap.poll();
            int second = minHeap.poll();
            int merged = first + second;
            totalCost += merged;
            minHeap.add(merged);
        }
        
        System.out.println("Total cost of merging: " + totalCost);
    }
}
            

Scenario:

Files of various sizes: 5, 25, 15, 30, 10, 20, 35.

Solution Approach:

Continue merging the smallest pairs using a priority queue to handle the larger dataset efficiently.

Code Example:

import java.util.PriorityQueue;

public class OptimalMergePattern {
    public static void main(String[] args) {
        int[] files = {5, 25, 15, 30, 10, 20, 35};
        PriorityQueue minHeap = new PriorityQueue<>();
        
        for (int file : files) {
            minHeap.add(file);
        }
        
        int totalCost = 0;
        while (minHeap.size() > 1) {
            int first = minHeap.poll();
            int second = minHeap.poll();
            int merged = first + second;
            totalCost += merged;
            minHeap.add(merged);
        }
        
        System.out.println("Total cost of merging: " + totalCost);
    }
}
            

Scenario:

Two large files of sizes 100 and 200 units.

Solution Approach:

In this case, the merge is straightforward since there are only two files. The total cost is simply the sum of their sizes.

Code Example:

import java.util.PriorityQueue;

public class OptimalMergePattern {
    public static void main(String[] args) {
        int[] files = {100, 200};
        PriorityQueue minHeap = new PriorityQueue<>();
        
        for (int file : files) {
            minHeap.add(file);
        }
        
        int totalCost = 0;
        while (minHeap.size() > 1) {
            int first = minHeap.poll();
            int second = minHeap.poll();
            int merged = first + second;
            totalCost += merged;
            minHeap.add(merged);
        }
        
        System.out.println("Total cost of merging: " + totalCost);
    }
}
            

Scenario:

Only one file of size 50 units.

Solution Approach:

No merge operation is needed as there is only one file. The cost is zero.

Code Example:

import java.util.PriorityQueue;

public class OptimalMergePattern {
    public static void main(String[] args) {
        int[] files = {50};
        PriorityQueue minHeap = new PriorityQueue<>();
        
        for (int file : files) {
            minHeap.add(file);
        }
        
        int totalCost = 0;
        while (minHeap.size() > 1) {
            int first = minHeap.poll();
            int second = minHeap.poll();
            int merged = first + second;
            totalCost += merged;
            minHeap.add(merged);
        }
        
        System.out.println("Total cost of merging: " + totalCost);
    }
}