Connect Ropes with Minimum Cost

Connect ropes with minimum cost using greedy approach with heap

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Connect Ropes with Minimum Cost

Problem Statement

You have some number of ropes with different lengths. You want to connect all ropes into one rope. The cost of connecting two ropes is equal to the sum of their lengths.

Return the minimum cost to connect all the ropes.

Input/Output Specifications

Input:
- ropes: Array of integers representing rope lengths
Output: Integer representing minimum cost to connect all ropes

Constraints

1 <= ropes.length <= 10^4
1 <= ropes[i] <= 10^4

Examples

Example 1:

Input: ropes = [8, 4, 6, 12]
Output: 58
Explanation: 
- Connect 4 and 6: cost = 10, remaining = [8, 10, 12]
- Connect 8 and 10: cost = 18, remaining = [12, 18]
- Connect 12 and 18: cost = 30, remaining = [30]
- Total cost = 10 + 18 + 30 = 58

Example 2:

Input: ropes = [20, 4, 8, 2]
Output: 54
Explanation:
- Connect 2 and 4: cost = 6, remaining = [6, 8, 20]
- Connect 6 and 8: cost = 14, remaining = [14, 20]
- Connect 14 and 20: cost = 34, remaining = [34]
- Total cost = 6 + 14 + 34 = 54

Example 3:

Input: ropes = [1, 2, 5, 10, 35, 89]
Output: 224
Explanation: Optimal strategy is to always connect two shortest ropes first.

Solution Approaches

Approach 1: Min-Heap (Optimal)

Algorithm Explanation:

Add all rope lengths to min-heap
While heap has more than one element:
- Extract two smallest ropes
- Connect them (cost = sum of lengths)
- Add total cost to result
- Put combined rope back to heap
Return total cost

Implementation:

Python:

import heapq

def min_cost_connect_ropes(ropes):
    """
    Connect ropes with minimum cost using min-heap
    Time: O(n log n)
    Space: O(n)
    """
    if len(ropes) <= 1:
        return 0
    
    # Create min-heap from rope lengths
    heap = ropes[:]
    heapq.heapify(heap)
    
    total_cost = 0
    
    # Connect ropes until only one remains
    while len(heap) > 1:
        # Get two shortest ropes
        first = heapq.heappop(heap)
        second = heapq.heappop(heap)
        
        # Cost to connect these ropes
        cost = first + second
        total_cost += cost
        
        # Put combined rope back
        heapq.heappush(heap, cost)
    
    return total_cost

# Alternative implementation with more explicit steps
def min_cost_connect_ropes_verbose(ropes):
    """
    Connect ropes with minimum cost (verbose version)
    Time: O(n log n)
    Space: O(n)
    """
    if not ropes or len(ropes) <= 1:
        return 0
    
    min_heap = []
    
    # Build min-heap
    for rope in ropes:
        heapq.heappush(min_heap, rope)
    
    total_cost = 0
    
    while len(min_heap) > 1:
        # Extract two minimum elements
        rope1 = heapq.heappop(min_heap)
        rope2 = heapq.heappop(min_heap)
        
        # Calculate cost and add to total
        current_cost = rope1 + rope2
        total_cost += current_cost
        
        # Insert merged rope back
        heapq.heappush(min_heap, current_cost)
    
    return total_cost

Java:

import java.util.*;

class Solution {
    /**
     * Connect ropes with minimum cost using min-heap
     * Time: O(n log n)
     * Space: O(n)
     */
    public int connectRopes(int[] ropes) {
        if (ropes.length <= 1) {
            return 0;
        }
        
        // Min-heap to store rope lengths
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        
        // Add all ropes to heap
        for (int rope : ropes) {
            minHeap.offer(rope);
        }
        
        int totalCost = 0;
        
        // Connect ropes until only one remains
        while (minHeap.size() > 1) {
            int first = minHeap.poll();
            int second = minHeap.poll();
            
            int cost = first + second;
            totalCost += cost;
            
            minHeap.offer(cost);
        }
        
        return totalCost;
    }
}

Go:

import (
    "container/heap"
)

type MinHeap []int

func (h MinHeap) Len() int           { return len(h) }
func (h MinHeap) Less(i, j int) bool { return h[i] < h[j] }
func (h MinHeap) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }

func (h *MinHeap) Push(x interface{}) {
    *h = append(*h, x.(int))
}

func (h *MinHeap) Pop() interface{} {
    old := *h
    n := len(old)
    x := old[n-1]
    *h = old[0 : n-1]
    return x
}

// connectRopes - Connect ropes with minimum cost using min-heap
// Time: O(n log n), Space: O(n)
func connectRopes(ropes []int) int {
    if len(ropes) <= 1 {
        return 0
    }
    
    // Create min-heap
    minHeap := &MinHeap{}
    heap.Init(minHeap)
    
    // Add all ropes to heap
    for _, rope := range ropes {
        heap.Push(minHeap, rope)
    }
    
    totalCost := 0
    
    // Connect ropes until only one remains
    for minHeap.Len() > 1 {
        first := heap.Pop(minHeap).(int)
        second := heap.Pop(minHeap).(int)
        
        cost := first + second
        totalCost += cost
        
        heap.Push(minHeap, cost)
    }
    
    return totalCost
}

JavaScript:

class MinHeap {
    constructor() {
        this.heap = [];
    }
    
    push(val) {
        this.heap.push(val);
        this.bubbleUp();
    }
    
    pop() {
        if (this.heap.length === 0) return null;
        if (this.heap.length === 1) return this.heap.pop();
        
        const min = this.heap[0];
        this.heap[0] = this.heap.pop();
        this.bubbleDown();
        return min;
    }
    
    size() {
        return this.heap.length;
    }
    
    bubbleUp() {
        let idx = this.heap.length - 1;
        while (idx > 0) {
            const parentIdx = Math.floor((idx - 1) / 2);
            if (this.heap[parentIdx] <= this.heap[idx]) break;
            [this.heap[parentIdx], this.heap[idx]] = [this.heap[idx], this.heap[parentIdx]];
            idx = parentIdx;
        }
    }
    
    bubbleDown() {
        let idx = 0;
        while (2 * idx + 1 < this.heap.length) {
            let minChildIdx = 2 * idx + 1;
            if (2 * idx + 2 < this.heap.length && 
                this.heap[2 * idx + 2] < this.heap[minChildIdx]) {
                minChildIdx = 2 * idx + 2;
            }
            
            if (this.heap[idx] <= this.heap[minChildIdx]) break;
            [this.heap[idx], this.heap[minChildIdx]] = [this.heap[minChildIdx], this.heap[idx]];
            idx = minChildIdx;
        }
    }
}

/**
 * Connect ropes with minimum cost using min-heap
 * Time: O(n log n)
 * Space: O(n)
 */
function connectRopes(ropes) {
    if (ropes.length <= 1) {
        return 0;
    }
    
    const minHeap = new MinHeap();
    
    // Add all ropes to heap
    for (const rope of ropes) {
        minHeap.push(rope);
    }
    
    let totalCost = 0;
    
    // Connect ropes until only one remains
    while (minHeap.size() > 1) {
        const first = minHeap.pop();
        const second = minHeap.pop();
        
        const cost = first + second;
        totalCost += cost;
        
        minHeap.push(cost);
    }
    
    return totalCost;
}

C#:

using System;
using System.Collections.Generic;

public class Solution {
    /// <summary>
    /// Connect ropes with minimum cost using min-heap
    /// Time: O(n log n)
    /// Space: O(n)
    /// </summary>
    public int ConnectRopes(int[] ropes) {
        if (ropes.Length <= 1) {
            return 0;
        }
        
        var minHeap = new PriorityQueue<int, int>();
        
        // Add all ropes to heap
        foreach (int rope in ropes) {
            minHeap.Enqueue(rope, rope);
        }
        
        int totalCost = 0;
        
        // Connect ropes until only one remains
        while (minHeap.Count > 1) {
            int first = minHeap.Dequeue();
            int second = minHeap.Dequeue();
            
            int cost = first + second;
            totalCost += cost;
            
            minHeap.Enqueue(cost, cost);
        }
        
        return totalCost;
    }
}

Approach 2: Sorting (Less Efficient)

Algorithm Explanation:

Sort ropes in ascending order
Always combine first two shortest ropes
Insert combined rope back in sorted order
Repeat until one rope remains

Implementation:

Python:

def min_cost_connect_ropes_sorting(ropes):
    """
    Connect ropes using sorting approach (less efficient)
    Time: O(n² log n) - due to repeated sorting
    Space: O(1)
    """
    if len(ropes) <= 1:
        return 0
    
    ropes = sorted(ropes)
    total_cost = 0
    
    while len(ropes) > 1:
        # Take two shortest ropes
        first = ropes.pop(0)
        second = ropes.pop(0)
        
        # Calculate cost
        cost = first + second
        total_cost += cost
        
        # Insert combined rope back in sorted order
        # Binary search for insertion point
        left, right = 0, len(ropes)
        while left < right:
            mid = (left + right) // 2
            if ropes[mid] <= cost:
                left = mid + 1
            else:
                right = mid
        
        ropes.insert(left, cost)
    
    return total_cost

Key Insights

Greedy Strategy: Always combine two shortest ropes for minimum cost
Huffman Coding: Similar to Huffman tree construction algorithm
Heap Efficiency: Min-heap provides O(log n) extraction and insertion
Cost Accumulation: Each connection adds to total cost

Edge Cases

Empty Array: No ropes to connect → Cost = 0
Single Rope: Already connected → Cost = 0
Two Ropes: Single connection → Cost = sum of both
All Same Length: Order doesn’t matter, but heap still optimal
Large Differences: Strategy remains same regardless of rope lengths

Test Cases

def test_min_cost_connect_ropes():
    # Example 1
    assert min_cost_connect_ropes([8, 4, 6, 12]) == 58
    
    # Example 2
    assert min_cost_connect_ropes([20, 4, 8, 2]) == 54
    
    # Single rope
    assert min_cost_connect_ropes([5]) == 0
    
    # Two ropes
    assert min_cost_connect_ropes([3, 7]) == 10
    
    # All same length
    assert min_cost_connect_ropes([5, 5, 5, 5]) == 30
    
    # Already optimal order
    assert min_cost_connect_ropes([1, 2, 3, 4]) == 19
    
    # Reverse order
    assert min_cost_connect_ropes([4, 3, 2, 1]) == 19
    
    print("All tests passed!")

test_min_cost_connect_ropes()

Interview Tips

Recognize Greedy Pattern: Always choosing two shortest ropes is optimal
Explain Why: Shorter ropes appear in more connections, so minimize their impact
Compare Approaches: Heap vs sorting trade-offs
Handle Edge Cases: Empty array, single rope scenarios
Trace Example: Walk through small example to show understanding

Follow-up Questions

Maximum Cost: What if we want maximum cost instead?
K-way Connection: Connect k ropes at once instead of 2
Memory Constraint: What if we can’t store all ropes in heap?
Streaming Input: Ropes arrive one by one
Different Costs: What if connection cost is not simple sum?

Common Mistakes

Wrong Strategy: Not always choosing shortest ropes first
Edge Cases: Not handling empty or single rope cases
Heap Operations: Incorrect heap push/pop sequence
Cost Calculation: Not accumulating total cost correctly
Termination: Wrong loop condition for heap size

Concept Explanations

Why Greedy Works: The optimal strategy is always to connect the two shortest ropes because shorter ropes will be part of more connections if we wait. By connecting them early, we minimize their contribution to the total cost.

Huffman Coding Connection: This problem is identical to constructing a Huffman tree for optimal prefix codes. The rope lengths correspond to character frequencies, and the connection costs correspond to the weighted path lengths in the tree.

Heap vs Sorting: While sorting might seem simpler, repeatedly finding and inserting minimum elements makes heap the natural choice. Heap operations are O(log n) vs O(n log n) for maintaining sorted order.

When to Apply: Use this pattern for optimization problems where you need to repeatedly select minimum elements and maintain a dynamic set. Common in greedy algorithms, scheduling problems, and tree construction algorithms.