Connect n ropes with minimum cost

Problem Statement

Given n ropes of different lengths, connect them into one rope with minimum cost. The cost of connecting two ropes is the sum of their lengths.

Example:

• Input: ropes = [4, 3, 2, 6]
• Output: 29
• Explanation:
  • Connect 2 and 3 → cost = 5, ropes = [4, 5, 6]
  • Connect 4 and 5 → cost = 9, ropes = [6, 9]
  • Connect 6 and 9 → cost = 15, ropes = [15]
  • Total cost = 5 + 9 + 15 = 29

Approach 1: Greedy without Heap (Brute Force)

Intuition

Always connect the two smallest ropes first. This minimizes the cost because smaller ropes contribute to subsequent connections fewer times.

Algorithm

1. Find two minimum length ropes
2. Connect them and add cost
3. Remove those two ropes, add the combined rope
4. Repeat until one rope remains

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class Solution {
    public int minCostToConnectRopes(int[] ropes) {
        if (ropes.length <= 1) return 0;
        
        List<Integer> ropeList = new ArrayList<>();
        for (int rope : ropes) {
            ropeList.add(rope);
        }
        
        int totalCost = 0;
        
        while (ropeList.size() > 1) {
            // Sort to find two minimum
            Collections.sort(ropeList);
            
            int first = ropeList.get(0);
            int second = ropeList.get(1);
            
            int cost = first + second;
            totalCost += cost;
            
            // Remove first two, add combined
            ropeList.remove(0);
            ropeList.remove(0);
            ropeList.add(cost);
        }
        
        return totalCost;
    }
}

Complexity Analysis

• Time Complexity: O(n² log n) - n-1 iterations, each with O(n log n) sorting
• Space Complexity: O(n) - For storing ropes

Approach 2: Min-Heap (Optimal)

Intuition

Use a min-heap to efficiently get the two smallest ropes. After combining, push the result back. Each operation is O(log n).

Algorithm

1. Build min-heap from all rope lengths
2. While heap has more than one element:
   • Extract two minimum elements
   • Combine them, add to cost
   • Push combined length back to heap
3. Return total cost

import java.util.PriorityQueue;

public class Solution {
    public int minCostToConnectRopes(int[] ropes) {
        if (ropes.length <= 1) return 0;
        
        // Min-heap
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        
        for (int rope : ropes) {
            minHeap.offer(rope);
        }
        
        int totalCost = 0;
        
        while (minHeap.size() > 1) {
            // Extract two minimum
            int first = minHeap.poll();
            int second = minHeap.poll();
            
            int cost = first + second;
            totalCost += cost;
            
            // Push combined back
            minHeap.offer(cost);
        }
        
        return totalCost;
    }
}

Complexity Analysis

• Time Complexity: O(n log n) - Building heap O(n), n-1 iterations with O(log n) operations each
• Space Complexity: O(n) - For the heap

Approach 3: Understanding the Greedy Choice

Intuition

Why does connecting smallest ropes first minimize cost? Each rope length contributes to the total cost based on how many times it's part of a connection. Smaller ropes should be connected first so they're part of fewer subsequent operations.

Mathematical Proof

Consider ropes with lengths a ≤ b ≤ c ≤ d.

Greedy (smallest first):

• Connect a + b = x, cost = x
• Connect x + c = y, cost = y = a + b + c
• Connect y + d = z, cost = z = a + b + c + d
• Total = (a+b) + (a+b+c) + (a+b+c+d) = 3a + 3b + 2c + d

Any other order will have larger coefficients for larger values.

import java.util.PriorityQueue;

public class Solution {
    public int minCostToConnectRopesExplained(int[] ropes) {
        if (ropes.length <= 1) return 0;
        
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        for (int rope : ropes) {
            minHeap.offer(rope);
        }
        
        int totalCost = 0;
        int step = 1;
        
        while (minHeap.size() > 1) {
            int first = minHeap.poll();
            int second = minHeap.poll();
            
            int cost = first + second;
            totalCost += cost;
            
            // System.out.println("Step " + step++ + ": Connect " + first + 
            //                    " + " + second + " = " + cost);
            
            minHeap.offer(cost);
        }
        
        return totalCost;
    }
}

Complexity Analysis

• Time Complexity: O(n log n)
• Space Complexity: O(n)

Related Problems

This problem is essentially the same as:

1. Huffman Coding - Build optimal prefix-free code
2. Minimum Cost to Merge Stones - Merge k stones at a time
3. Minimum Cost Tree from Leaf Values - Similar tree building concept

Complexity Analysis

• Time Complexity: O(n log n)
• Space Complexity: O(n)

Key Takeaways

1. Greedy approach: Always connect smallest ropes first
2. Min-heap enables efficient O(log n) extraction of minimum
3. This is equivalent to Huffman Coding problem
4. Each element contributes to cost based on its "depth" in merge tree
5. Total n-1 merge operations needed for n ropes
6. The problem teaches the importance of choosing data structures wisely
7. Brute force sorting is O(n²), heap approach is O(n log n)
8. This is LeetCode 1167 (Minimum Cost to Connect Sticks)