Efficiently Connecting N Ropes with Minimum Cost in Java: A Step-by-Step Guide

 

Connecting N Ropes with Minimum Cost in Java

In this blog post, we’ll explore a Java program that connects N ropes with the minimum cost. The problem is a classic example of using a priority queue (or min-heap) to efficiently solve a problem that involves repeatedly combining the smallest elements.

Code Explanation

Here’s the Java code for the problem:

import java.util.*;

public class QueueConnectNRopes {
    public static int minPrice(int arr[]){
        Queue <Integer> q = new LinkedList<>();

        for (int i = 0; i < arr.length; i++) {
            q.add(arr[i]);
        }
        
        int total = 0;
        while (!q.isEmpty()) {
            int cost = q.remove() + q.remove();
            q.add(cost);
            total += cost;   
            if (q.size() == 1) {
                return total;
            }
        }
        return total;
    }
    public static void main(String[] args) {
        int arr[] = {1,2,3,4};

        System.out.println(minPrice(arr));
    }
}

How It Works

1. Initialization: We initialize a queue and add all elements of the array to it.
2. Combining Ropes: We repeatedly remove the two smallest elements from the queue, combine them, and add the combined length back to the queue. The cost of each combination is added to the total cost.
3. Termination: The process continues until only one element remains in the queue, which represents the total cost of combining all ropes.

Time Complexity

The time complexity of this algorithm is O(n log n). Here’s why:

• Adding all elements to the queue takes O(n) time.
• Each combination operation involves removing two elements and adding one element back to the queue. Since the queue is implemented as a min-heap, each remove and add operation takes O(log n) time.
• Since we perform this operation n-1 times, the total time complexity is O(n log n).

Space Complexity

The space complexity of this algorithm is O(n). This is because we store all the elements in the queue, which requires O(n) space.

Conclusion

This Java program efficiently solves the problem of connecting N ropes with the minimum cost using a priority queue. The use of a min-heap ensures that the smallest elements are always combined first, leading to an optimal solution. The time complexity of O(n log n) and space complexity of O(n) make this approach both efficient and scalable.

Feel free to ask if you have any questions or need further clarification on any part of the code!