Minimum Cost of ropes

Problem Statement

Given n ropes of different lengths, connect all ropes into one rope. The cost to connect two ropes is equal to the sum of their lengths. Find the minimum cost to connect all ropes.

Constraints:

• 1 ≤ n ≤ 10^5
• 1 ≤ rope[i] ≤ 10^6

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 = [9, 6]
  • Connect 6 and 9 → cost = 15, ropes = [15]
  • Total = 5 + 9 + 15 = 29

Example 2:

• Input: ropes = [1, 2, 3]
• Output: 9
• Explanation: Connect 1+2=3 (cost 3), then 3+3=6 (cost 6). Total = 9

Approach 1: Brute Force (Always Find Minimum)

Intuition

At each step, find the two shortest ropes and connect them. This requires scanning the array each time to find the minimum elements.

Algorithm

1. While more than one rope exists:
   • Find two shortest ropes
   • Connect them (add to cost)
   • Replace them with the connected rope
2. Return total cost

import java.util.*;

public class Solution {
    public long minCostBrute(int[] ropes) {
        List<Integer> remaining = new ArrayList<>();
        for (int r : ropes) remaining.add(r);
        
        long totalCost = 0;
        
        while (remaining.size() > 1) {
            Collections.sort(remaining);
            
            int first = remaining.get(0);
            int second = remaining.get(1);
            int combined = first + second;
            
            totalCost += combined;
            
            remaining.remove(0);
            remaining.remove(0);
            remaining.add(combined);
        }
        
        return totalCost;
    }
}

Complexity Analysis

• Time Complexity: O(n²) - n iterations, each with O(n) or O(n log n) operations
• Space Complexity: O(n) - For storing remaining ropes

Approach 2: Min-Heap / Priority Queue (Optimal)

Intuition

Use a min-heap to efficiently extract the two smallest ropes. The heap maintains the invariant that we can always get the minimum in O(log n) time.

Greedy Insight: Connecting smaller ropes first minimizes cost because smaller ropes contribute to multiple future connections (their length gets added multiple times).

Algorithm

1. Build a min-heap from all rope lengths
2. While heap has more than one element:
   • Extract two minimum ropes
   • Connect them (add to cost)
   • Insert the combined rope back
3. Return total cost

import java.util.*;

public class Solution {
    public long minCostRopes(int[] ropes) {
        // Min-heap
        PriorityQueue<Long> pq = new PriorityQueue<>();
        
        // Insert all ropes
        for (int rope : ropes) {
            pq.offer((long)rope);
        }
        
        long totalCost = 0;
        
        // Connect ropes until only one remains
        while (pq.size() > 1) {
            long first = pq.poll();
            long second = pq.poll();
            
            long combined = first + second;
            totalCost += combined;
            
            pq.offer(combined);
        }
        
        return totalCost;
    }
    
    public static void main(String[] args) {
        Solution sol = new Solution();
        int[] ropes = {4, 3, 2, 6};
        System.out.println(sol.minCostRopes(ropes));  // Output: 29
    }
}

Dry Run Example

Why Greedy Works - Proof

Consider any rope of length L. It contributes to the total cost based on its depth in the connection tree.

If rope L is at depth d in the tree:

• It gets included in d connections
• Total contribution = L × d

Observation: Shorter ropes should be connected earlier (deeper in tree) because their small length is multiplied many times.

This is exactly what our greedy algorithm achieves - smallest ropes get connected first, placing them deepest in the tree.

Complexity Analysis

• Time Complexity: O(n log n) - Building heap is O(n), then (n-1) extract-min and insert operations, each O(log n)
• Space Complexity: O(n) - For the heap

Huffman Coding Connection

This problem is equivalent to Huffman Coding where:

• Rope lengths = Character frequencies
• Connection cost = Building Huffman tree cost

The optimal prefix-free code (Huffman) minimizes the weighted path length, which corresponds to minimizing the total rope connection cost.

Key Takeaways

1. Greedy Strategy: Always connect the two shortest ropes
2. Min-Heap: Essential for efficient minimum extraction
3. Huffman Connection: Same algorithm as Huffman coding
4. Tree Depth Matters: Shorter ropes should be deeper in connection tree
5. Time Optimization: Heap reduces O(n²) brute force to O(n log n)
6. Practical Applications: File merging, data compression, optimal merge patterns
7. Integer Overflow: Use long long for large inputs