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Connect Ropes with Minimum Cost

A curated StudyFlow practice problem with comprehensive editorial and multi-language solutions.

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Problem Statement

There are N ropes of given lengths in an array arr. We need to connect these ropes into one rope. We can connect only two ropes at a time. The cost of connecting two ropes of lengths A and B is A + B.

Find the minimum possible total cost to connect all ropes.

Example 1 · Example 1: Four Ropes:

Input: arr = [4, 3, 2, 6]
Output: 29
Explanation: 1. Connect ropes of lengths 2 and 3: cost = 2 + 3 = 5. Ropes left: [4, 6, 5]. Total cost = 5. 2. Connect ropes of lengths 4 and 5: cost = 4 + 5 = 9. Ropes left: [6, 9]. Total cost = 5 + 9 = 14. 3. Connect ropes of lengths 6 and 9: cost = 6 + 9 = 15. Ropes left: [15]. Total cost = 14 + 15 = 29. Merge Tree: (29) / \ (6) (9) / \ (4) (5) / \ (2) (3)

Constraints:

  • 1 <= N <= 105
  • 1 <= arr[i] <= 106

What You'll Learn

Key principles and algorithmic techniques you will master from this problem:

  • Recognize the greedy Huffman property in rope merging.
  • Implement a priority queue solution across 4 languages.
  • Understand the two-queue optimization technique.

Pattern Recognition Trigger

How to instantly recognize this algorithmic pattern when encountering unseen interview questions:

Signals to Look For:

  • Repeatedly combine two smallest entities and pay their sum as cost.
  • Greedy minimization of cumulative intermediate sums.
  • Identical to Huffman coding tree construction.

Algorithmic Decision Tree:

  • Use Min-Heap (PriorityQueue) to dynamically keep track of smallest ropes.
  • Extract 2 elements, sum them, add to total cost, and re-insert the sum.

Connected Patterns:

Minimum Cost to Connect SticksMerge k Sorted Lists

1. Brute Force — Linear Scan Merge

Intuition

Loop N-1 times. Find the minimum element, remove it, find the second minimum, remove it, add their sum to total cost, and append the sum.

When to apply: When external libraries are unavailable and N is small (N <= 100).

Algorithm

  1. Step 1: Check if size <= 1; return 0.
  2. Step 2: Repeat N-1 times: find min1, remove; find min2, remove.
  3. Step 3: cost += min1 + min2.
  4. Step 4: Append min1 + min2.
  5. Step 5: Return total cost.
C++ SOLUTIONCLEAN & OPTIMIZED
1#include <iostream>
2#include <vector>
3#include <algorithm>
4using namespace std;
5
6class Solution {
7public:
8 long long minCost(vector<long long>& arr) {
9 if (arr.size() <= 1) return 0;
10 long long total = 0;
11 while (arr.size() > 1) {
12 auto it1 = min_element(arr.begin(), arr.end());
13 long long a = *it1;
14 arr.erase(it1);
15
16 auto it2 = min_element(arr.begin(), arr.end());
17 long long b = *it2;
18 arr.erase(it2);
19
20 long long sum = a + b;
21 total += sum;
22 arr.push_back(sum);
23 }
24 return total;
25 }
26};
27
28int main() {
29 Solution sol;
30 vector<long long> arr = {4, 3, 2, 6};
31 cout << "Min Cost (Brute Force): " << sol.minCost(arr) << endl;
32 return 0;
33}

Time & Space Complexity

  • Time complexity: O(N^2) — O(N2) due to repeated linear scans.
  • Space complexity: O(N) — O(N) to store dynamic rope list.

Why Brute Force Doesn't Scale:

For N = 105, O(N2) takes ~1010 operations, whereas Min-Heap takes ~1.7 * 106 operations.

Min-Heap completes in ~20ms for N = 100,000 ropes.

2. Better — Two Queues

Intuition

By pre-sorting, original ropes are in order. Merged sums are also created in monotonically non-decreasing order. Thus, comparing front of both queues takes O(1).

When to apply: When sorting can be done once and fast linear merges are preferred.

Algorithm

  1. Step 1: Sort ropes in ascending order.
  2. Step 2: Push all into q1; initialize empty q2.
  3. Step 3: While q1.size + q2.size > 1, extract two minimums from front of queues.
  4. Step 4: Sum them, add to total cost, and push sum into q2.
  5. Step 5: Return total cost.
C++ SOLUTIONCLEAN & OPTIMIZED
1#include <iostream>
2#include <vector>
3#include <queue>
4#include <algorithm>
5using namespace std;
6
7class Solution {
8public:
9 long long minCost(vector<long long>& arr) {
10 if (arr.size() <= 1) return 0;
11 sort(arr.begin(), arr.end());
12 queue<long long> q1, q2;
13 for (long long x : arr) q1.push(x);
14
15 long long total = 0;
16 auto getMin = [&]() {
17 if (q1.empty()) { long long v = q2.front(); q2.pop(); return v; }
18 if (q2.empty()) { long long v = q1.front(); q1.pop(); return v; }
19 if (q1.front() <= q2.front()) { long long v = q1.front(); q1.pop(); return v; }
20 long long v = q2.front(); q2.pop(); return v;
21 };
22
23 while (q1.size() + q2.size() > 1) {
24 long long a = getMin();
25 long long b = getMin();
26 long long sum = a + b;
27 total += sum;
28 q2.push(sum);
29 }
30 return total;
31 }
32};
33
34int main() {
35 Solution sol;
36 vector<long long> arr = {4, 3, 2, 6};
37 cout << "Min Cost (Two Queues): " << sol.minCost(arr) << endl;
38 return 0;
39}

Time & Space Complexity

  • Time complexity: O(N log N) — O(N log N) from the initial array sort.
  • Space complexity: O(N) — O(N) for queue allocations.

3. Optimal — Min-Heap (Priority Queue)

Intuition

Using a binary min-heap guarantees that finding and extracting the smallest elements takes O(log N). Performing this N-1 times yields O(N log N) total runtime.

When to apply: Standard optimal greedy algorithm.

Algorithm

  1. Step 1: If rope count <= 1, return 0.
  2. Step 2: Insert all rope lengths into a Min-Heap.
  3. Step 3: While heap size > 1, extract smallest rope a and second smallest b.
  4. Step 4: Sum = a + b. Add sum to total cost.
  5. Step 5: Push sum back into Min-Heap.
  6. Step 6: Return total cost.

Interactive Trace for Connect Ropes with Minimum Cost

2 of 2 questions left today
Step 1 / 3
Trace timeline33% explored
Current operationSTEP 1 OF 3
Step 1: Heap initialized with [2, 3, 4, 6]
Visual state
Values and pointers at this exact step
tree
Root: 2
2
Child: 3
3
Child: 4
4
Leaf: 6
6

Node order shown; add connections to display explicit tree edges.

merge: r1
Check / comparison
sum = 2 + 3 = 5, cost = 5
Decision / next transition
Push 5 into heap. Heap becomes [4, 5, 6].
Why this step matters

Step 1: Heap initialized with [2, 3, 4, 6]. Extract two smallest: 2 and 3.

Step 1 / 3
C++ SOLUTIONCLEAN & OPTIMIZED
1#include <iostream>
2#include <vector>
3#include <queue>
4using namespace std;
5
6class Solution {
7public:
8 long long minCost(vector<long long>& arr) {
9 if (arr.size() <= 1) return 0;
10 priority_queue<long long, vector<long long>, greater<long long>> minHeap(arr.begin(), arr.end());
11
12 long long total = 0;
13 while (minHeap.size() > 1) {
14 long long a = minHeap.top(); minHeap.pop();
15 long long b = minHeap.top(); minHeap.pop();
16 long long sum = a + b;
17 total += sum;
18 minHeap.push(sum);
19 }
20 return total;
21 }
22};
23
24int main() {
25 Solution sol;
26 vector<long long> arr = {4, 3, 2, 6};
27 cout << "Min Cost (Optimal Min-Heap): " << sol.minCost(arr) << endl;
28 return 0;
29}

Time & Space Complexity

  • Time complexity: O(N log N) — O(N log N) heap processing.
  • Space complexity: O(N) — O(N) heap storage.

Approach Comparison Matrix

Side-by-side comparison of time, space, and algorithmic trade-offs across each approach:

ApproachTimeSpaceCore IdeaTrade-off / Bottleneck
1.Brute Force — Linear Scan & Array ModificationO(N^2)O(1) auxiliaryScan array N-1 times to find two smallest ropes, merge them and erase one element.Repeated linear minimum searches.
2.Better — Two QueuesO(N log N)O(N)Sort original ropes into Queue 1. Store merged ropes in Queue 2. Greedily pick smallest from heads.Initial sorting.
3.Optimal — Min-Heap Priority QueueOptimalO(N log N)O(N)Insert all rope lengths into Min-Heap. Repeatedly extract two smallest ropes and insert their sum.None. Optimal.

Common Pitfalls & Crucial Edge Cases

Using 32-bit integer

Sums easily overflow for N = 105 and lengths = 106.

Accumulating lengths instead of costs

Cost paid is the sum at each step, not just the final rope length.

Not re-inserting combined rope

The combined rope must participate in subsequent merges.

Crucial Edge Cases to Validate:

  • Single rope (N = 1) — Return 0, no connections needed.
  • Two ropes (N = 2) — Return sum of the two ropes.
  • All identical lengths — Handled naturally by min-heap order.
  • Integer overflow — Use 64-bit integers (long / long long) as cumulative sums can exceed 231 - 1.

Interview Tips & Best Practices

How to communicate your thought process and address follow-ups in a live technical interview:

  1. Clarify integer overflow immediately: mention using long long in C++ and long in Java.
  2. Mention Huffman coding.

Quick Comprehension Check

Test your intuitive understanding of the core concept in 30 seconds:

1.What mathematical property justifies merging the two smallest ropes first?

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