Connect Ropes (easy)

Problem Statement

Given ‘N’ ropes with different lengths, we need to connect these ropes into one big rope with minimum cost. The cost of connecting two ropes is equal to the sum of their lengths.

Example 1:

Input: [1, 3, 11, 5]
Output: 33
Explanation: First connect 1+3(=4), then 4+5(=9), and then 9+11(=20). So the total cost is 33 (4+9+20)

Example 2:

Input: [3, 4, 5, 6]
Output: 36
Explanation: First connect 3+4(=7), then 5+6(=11), 7+11(=18). Total cost is 36 (7+11+18)

Example 3:

Input: [1, 3, 11, 5, 2]
Output: 42
Explanation: First connect 1+2(=3), then 3+3(=6), 6+5(=11), 11+11(=22). Total cost is 42 (3+6+11+22)

Try it yourself

Try solving this question here:

import java.util.*;
class ConnectRopes {
  public static int minimumCostToConnectRopes(int[] ropeLengths) {
    // TODO: Write your code here
    return -1;
  }
  public static void main(String[] args) {
    int result = ConnectRopes.minimumCostToConnectRopes(new int[] { 1, 3, 11, 5 });
    System.out.println("Minimum cost to connect ropes: " + result);
    result = ConnectRopes.minimumCostToConnectRopes(new int[] { 3, 4, 5, 6 });
    System.out.println("Minimum cost to connect ropes: " + result);
    result = ConnectRopes.minimumCostToConnectRopes(new int[] { 1, 3, 11, 5, 2 });
    System.out.println("Minimum cost to connect ropes: " + result);
  }
}

Introduction

Pattern: Sliding Window

Pattern: Two Pointers

Pattern: Fast and Slow Patterns

Pattern: Merge Intervals

Pattern: Cyclic Sort

Pattern: In-place Reversal of a LinkedList

Pattern: Tree Breadth First Search

Pattern: Tree Depth First Search

Pattern: Two Heaps

Pattern: Subsets

Pattern: Modified Binary Search

Pattern: Bitwise XOR

Pattern: Top 'K' Elements

Pattern: K-way merge

Pattern : 0/1 Knapsack (Dynamic Programming)

Pattern: Topological Sort (Graph)

Miscellaneous

Conclusions

Connect Ropes (easy)

Problem Statement

Try it yourself

Solution