Connect n ropes with minimum cost

Given are N ropes of different lengths, the task is to connect these ropes into one rope with minimum cost, such that the cost to connect two ropes is equal to the sum of their lengths.

Examples:

Input: arr[] = {4,3,2,6} , N = 4
Output: 29
Explanation:

1. First, connect ropes of lengths 2 and 3. Now we have three ropes of lengths 4, 6, and 5. 
2. Now connect ropes of lengths 4 and 5. Now we have two ropes of lengths 6 and 9. 
3. Finally connect the two ropes and all ropes have connected.

Input: arr[] = {1, 2, 3} , N = 3
Output: 9
Explanation:

1. First, connect ropes of lengths 1 and 2. Now we have two ropes of lengths 3 and 3. 
2. Finally connect the two ropes and all ropes have connected.

We strongly recommend that you click here and practice it, before moving on to the solution.

Connect N ropes with minimum cost using Min-Heap

Approach: If we observe the above problem closely, we can notice that the lengths of the ropes which are picked first are included more than once in the total cost. Therefore, the idea is to connect the smallest two ropes first and recur for the remaining ropes. This approach is similar to Huffman Coding. We put the smallest ropes down the tree so they can be repeated multiple times rather than the longer ones.

Illustration:

• First, we will connect ropes of lengths 2 and 3 because they are the smallest. Now we have three ropes left of lengths 4, 6, and 5.
• Now we connect ropes of lengths 4 and 5. Now we have two ropes of lengths 6 and 9.
• Finally, we will connect the two ropes so that all ropes are connected.

• The total cost contains the sum of depth of each value. For array [ 2, 3, 4, 6 ] the sum is equal to (2 * 3) + (3 * 3) + (4 * 2) + (6 * 1) = 29 (According to the diagram). 

Algorithm: Follow the steps mentioned below to implement the idea:

• Create a min-heap and insert all lengths into the min-heap.
• Do following while the number of elements in min-heap is greater than one. 
  • Extract the minimum and second minimum from min-heap
  • Add the above two extracted values and insert the added value to the min-heap.
  • Maintain a variable for total cost and keep incrementing it by the sum of extracted values.
• Return the value of total cost.

Below is the implementation of the above approach:

C++

// C++ program for connecting
// n ropes with minimum cost
#include <bits/stdc++.h>

using namespace std;

// A Min Heap: Collection of min heap nodes
struct MinHeap {
    unsigned size; // Current size of min heap
    unsigned capacity; // capacity of min heap
    int* harr; // Array of minheap nodes
};

// A utility function to create
// a min-heap of a given capacity
struct MinHeap* createMinHeap(unsigned capacity)
{
    struct MinHeap* minHeap = new MinHeap;
    minHeap->size = 0; // current size is 0
    minHeap->capacity = capacity;
    minHeap->harr = new int[capacity];
    return minHeap;
}

// A utility function to swap two min heap nodes
void swapMinHeapNode(int* a, int* b)
{
    int temp = *a;
    *a = *b;
    *b = temp;
}

// The standard minHeapify function.
void minHeapify(struct MinHeap* minHeap, int idx)
{
    int smallest = idx;
    int left = 2 * idx + 1;
    int right = 2 * idx + 2;

    if (left < minHeap->size
        && minHeap->harr[left] < minHeap->harr[smallest])
        smallest = left;

    if (right < minHeap->size
        && minHeap->harr[right] < minHeap->harr[smallest])
        smallest = right;

    if (smallest != idx) {
        swapMinHeapNode(&minHeap->harr[smallest],
                        &minHeap->harr[idx]);
        minHeapify(minHeap, smallest);
    }
}

// A utility function to check
// if size of heap is 1 or not
int isSizeOne(struct MinHeap* minHeap)
{
    return (minHeap->size == 1);
}

// A standard function to extract
// minimum value node from heap
int extractMin(struct MinHeap* minHeap)
{
    int temp = minHeap->harr[0];
    minHeap->harr[0] = minHeap->harr[minHeap->size - 1];
    --minHeap->size;
    minHeapify(minHeap, 0);
    return temp;
}

// A utility function to insert
// a new node to Min Heap
void insertMinHeap(struct MinHeap* minHeap, int val)
{
    ++minHeap->size;
    int i = minHeap->size - 1;
    while (i && (val < minHeap->harr[(i - 1) / 2])) {
        minHeap->harr[i] = minHeap->harr[(i - 1) / 2];
        i = (i - 1) / 2;
    }
    minHeap->harr[i] = val;
}

// A standard function to build min-heap
void buildMinHeap(struct MinHeap* minHeap)
{
    int n = minHeap->size - 1;
    int i;
    for (i = (n - 1) / 2; i >= 0; --i)
        minHeapify(minHeap, i);
}

// Creates a min-heap of capacity
// equal to size and inserts all values
// from len[] in it. Initially, size
// of min heap is equal to capacity
struct MinHeap* createAndBuildMinHeap(int len[], int size)
{
    struct MinHeap* minHeap = createMinHeap(size);
    for (int i = 0; i < size; ++i)
        minHeap->harr[i] = len[i];
    minHeap->size = size;
    buildMinHeap(minHeap);
    return minHeap;
}

// The main function that returns
// the minimum cost to connect n
// ropes of lengths stored in len[0..n-1]
int minCost(int len[], int n)
{
    int cost = 0; // Initialize result

    // Create a min heap of capacity
    // equal to n and put all ropes in it
    struct MinHeap* minHeap = createAndBuildMinHeap(len, n);

    // Iterate while size of heap doesn't become 1
    while (!isSizeOne(minHeap)) {
        // Extract two minimum length
        // ropes from min heap
        int min = extractMin(minHeap);
        int sec_min = extractMin(minHeap);

        cost += (min + sec_min); // Update total cost

        // Insert a new rope in min heap
        // with length equal to sum
        // of two extracted minimum lengths
        insertMinHeap(minHeap, min + sec_min);
    }

    // Finally return total minimum
    // cost for connecting all ropes
    return cost;
}

// Driver program to test above functions
int main()
{
    int len[] = { 4, 3, 2, 6 };
    int size = sizeof(len) / sizeof(len[0]);
    cout << "Total cost for connecting ropes is "
         << minCost(len, size);
    return 0;
}

Java

// Java program to connect n
// ropes with minimum cost

// A class for Min Heap
class MinHeap {
    int[] harr; // Array of elements in heap
    int heap_size; // Current number of elements in min heap
    int capacity; // maximum possible size of min heap

    // Constructor: Builds a heap from
    // a given array a[] of given size
    public MinHeap(int a[], int size)
    {
        heap_size = size;
        capacity = size;
        harr = a;
        int i = (heap_size - 1) / 2;
        while (i >= 0) {
            MinHeapify(i);
            i--;
        }
    }

    // A recursive method to heapify a subtree
    // with the root at given index
    // This method assumes that the subtrees
    // are already heapified
    void MinHeapify(int i)
    {
        int l = left(i);
        int r = right(i);
        int smallest = i;
        if (l < heap_size && harr[l] < harr[i])
            smallest = l;
        if (r < heap_size && harr[r] < harr[smallest])
            smallest = r;
        if (smallest != i) {
            swap(i, smallest);
            MinHeapify(smallest);
        }
    }

    int parent(int i) { return (i - 1) / 2; }

    // to get index of left child of node at index i
    int left(int i) { return (2 * i + 1); }

    // to get index of right child of node at index i
    int right(int i) { return (2 * i + 2); }

    // Method to remove minimum element (or root) from min
    // heap
    int extractMin()
    {
        if (heap_size <= 0)
            return Integer.MAX_VALUE;
        if (heap_size == 1) {
            heap_size--;
            return harr[0];
        }

        // Store the minimum value, and remove it from heap
        int root = harr[0];
        harr[0] = harr[heap_size - 1];
        heap_size--;
        MinHeapify(0);

        return root;
    }

    // Inserts a new key 'k'
    void insertKey(int k)
    {
        if (heap_size == capacity) {
            System.out.println(
                "Overflow: Could not insertKey");
            return;
        }

        // First insert the new key at the end
        heap_size++;
        int i = heap_size - 1;
        harr[i] = k;

        // Fix the min heap property if it is violated
        while (i != 0 && harr[parent(i)] > harr[i]) {
            swap(i, parent(i));
            i = parent(i);
        }
    }

    // A utility function to check
    // if size of heap is 1 or not
    boolean isSizeOne() { return (heap_size == 1); }

    // A utility function to swap two elements
    void swap(int x, int y)
    {
        int temp = harr[x];
        harr[x] = harr[y];
        harr[y] = temp;
    }

    // The main function that returns the
    // minimum cost to connect n ropes of
    // lengths stored in len[0..n-1]
    static int minCost(int len[], int n)
    {
        int cost = 0; // Initialize result

        // Create a min heap of capacity equal
        // to n and put all ropes in it
        MinHeap minHeap = new MinHeap(len, n);

        // Iterate while size of heap doesn't become 1
        while (!minHeap.isSizeOne()) {
            // Extract two minimum length ropes from min
            // heap
            int min = minHeap.extractMin();
            int sec_min = minHeap.extractMin();

            cost += (min + sec_min); // Update total cost

            // Insert a new rope in min heap with length
            // equal to sum of two extracted minimum lengths
            minHeap.insertKey(min + sec_min);
        }

        // Finally return total minimum
        // cost for connecting all ropes
        return cost;
    }

    // Driver code
    public static void main(String args[])
    {
        int len[] = { 4, 3, 2, 6 };
        int size = len.length;

        System.out.println(
            "Total cost for connecting ropes is "
            + minCost(len, size));
    }
};

// This code is contributed by shubham96301

C#

// C# program to connect n ropes with minimum cost
using System;

// A class for Min Heap
class MinHeap {
    int[] harr; // Array of elements in heap
    int heap_size; // Current number of elements in min heap
    int capacity; // maximum possible size of min heap

    // Constructor: Builds a heap from
    // a given array a[] of given size
    public MinHeap(int[] a, int size)
    {
        heap_size = size;
        capacity = size;
        harr = a;
        int i = (heap_size - 1) / 2;
        while (i >= 0) {
            MinHeapify(i);
            i--;
        }
    }

    // A recursive method to heapify a subtree
    // with the root at given index
    // This method assumes that the subtrees
    // are already heapified
    void MinHeapify(int i)
    {
        int l = left(i);
        int r = right(i);
        int smallest = i;
        if (l < heap_size && harr[l] < harr[i])
            smallest = l;
        if (r < heap_size && harr[r] < harr[smallest])
            smallest = r;
        if (smallest != i) {
            swap(i, smallest);
            MinHeapify(smallest);
        }
    }

    int parent(int i) { return (i - 1) / 2; }

    // to get index of left child of node at index i
    int left(int i) { return (2 * i + 1); }

    // to get index of right child of node at index i
    int right(int i) { return (2 * i + 2); }

    // Method to remove minimum element (or root) from min
    // heap
    int extractMin()
    {
        if (heap_size <= 0)
            return int.MaxValue;
        if (heap_size == 1) {
            heap_size--;
            return harr[0];
        }

        // Store the minimum value, and remove it from heap
        int root = harr[0];
        harr[0] = harr[heap_size - 1];
        heap_size--;
        MinHeapify(0);

        return root;
    }

    // Inserts a new key 'k'
    void insertKey(int k)
    {
        if (heap_size == capacity) {
            Console.WriteLine(
                "Overflow: Could not insertKey");
            return;
        }

        // First insert the new key at the end
        heap_size++;
        int i = heap_size - 1;
        harr[i] = k;

        // Fix the min heap property if it is violated
        while (i != 0 && harr[parent(i)] > harr[i]) {
            swap(i, parent(i));
            i = parent(i);
        }
    }

    // A utility function to check
    // if size of heap is 1 or not
    Boolean isSizeOne() { return (heap_size == 1); }

    // A utility function to swap two elements
    void swap(int x, int y)
    {
        int temp = harr[x];
        harr[x] = harr[y];
        harr[y] = temp;
    }

    // The main function that returns the
    // minimum cost to connect n ropes of
    // lengths stored in len[0..n-1]
    static int minCost(int[] len, int n)
    {
        int cost = 0; // Initialize result

        // Create a min heap of capacity equal
        // to n and put all ropes in it
        MinHeap minHeap = new MinHeap(len, n);

        // Iterate while size of heap doesn't become 1
        while (!minHeap.isSizeOne()) {
            // Extract two minimum length ropes from min
            // heap
            int min = minHeap.extractMin();
            int sec_min = minHeap.extractMin();

            cost += (min + sec_min); // Update total cost

            // Insert a new rope in min heap with length
            // equal to sum of two extracted minimum lengths
            minHeap.insertKey(min + sec_min);
        }

        // Finally return total minimum
        // cost for connecting all ropes
        return cost;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int[] len = { 4, 3, 2, 6 };
        int size = len.Length;

        Console.WriteLine(
            "Total cost for connecting ropes is "
            + minCost(len, size));
    }
};

// This code is contributed by Arnab Kundu

Python3

# Python3 program to connect n
# ropes with minimum cost
import heapq


def minCost(arr, n):

    # Create a priority queue out of the
    # given list
    heapq.heapify(arr)

    # Initialize result
    res = 0

    # While size of priority queue
    # is more than 1
    while(len(arr) > 1):

        # Extract shortest two ropes from arr
        first = heapq.heappop(arr)
        second = heapq.heappop(arr)

        # Connect the ropes: update result
        # and insert the new rope to arr
        res += first + second
        heapq.heappush(arr, first + second)

    return res


# Driver code
if __name__ == '__main__':

    lengths = [4, 3, 2, 6]
    size = len(lengths)

    print("Total cost for connecting ropes is " +
          str(minCost(lengths, size)))

# This code is contributed by shivampatel5

Javascript

<script>

// JavaScript program to connect n
// ropes with minimum cost

function minCost(arr,n)
{
    // Create a priority queue
        let pq = [];

        // Adding items to the pQueue
        for (let i = 0; i < n; i++) {
            pq.push(arr[i]);
        }

        pq.sort(function(a,b){return a-b;});

        // Initialize result
        let res = 0;

        // While size of priority queue
        // is more than 1
        while (pq.length > 1) {
            // Extract shortest two ropes from pq
            let first = pq.shift();
            let second = pq.shift();

            // Connect the ropes: update result
            // and insert the new rope to pq
            res += first + second;
            pq.push(first + second);
            pq.sort(function(a,b){return a-b;});
        }

        return res;
}

// Driver program to test above function
let len = [4, 3, 2, 6];
let size = len.length;
document.write("Total cost for connecting"
                           + " ropes is " + minCost(len, size));

// This code is contributed by avanitrachhadiya2155

</script>

Output

Total cost for connecting ropes is 29

Time Complexity: O(N*log(N))
Auxiliary Space: O(N)

Connect N ropes with minimum cost using Pre Defined Function

In this approach, we use the predefined priority queue which is already available. The approach and algorithm remain the same. The min heap is replaced by a priority queue.

Follow the steps mentioned below to implement the idea:

1. declare a priority queue and push all the elements in it.
2. Do following while the number of elements in min-heap is greater than one. 
   1. Extract the minimum and second minimum from min-heap
   2. Add the above two extracted values and insert the added value to the min-heap.
   3. Maintain a variable for total cost and keep incrementing it by the sum of extracted values.
3. Return the value of total cost.

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>

using namespace std;

int minCost(int arr[], int n)
{
    // Create a priority queue
    // https://www.geeksforgeeks.org/priority-queue-in-cpp-stl/
    // By default 'less' is used which is for decreasing
    // order and 'greater' is used for increasing order
    priority_queue<int, vector<int>, greater<int> > pq(
        arr, arr + n);

    // Initialize result
    int res = 0;

    // While size of priority queue is more than 1
    while (pq.size() > 1) {
        // Extract shortest two ropes from pq
        int first = pq.top();
        pq.pop();
        int second = pq.top();
        pq.pop();

        // Connect the ropes: update result and
        // insert the new rope to pq
        res += first + second;
        pq.push(first + second);
    }

    return res;
}

// Driver program to test above function
int main()
{
    int len[] = { 4, 3, 2, 6 };
    int size = sizeof(len) / sizeof(len[0]);
    cout << "Total cost for connecting ropes is "
         << minCost(len, size);
    return 0;
}

Java

// Java program to connect n
// ropes with minimum cost
import java.util.*;

class ConnectRopes {
    static int minCost(int arr[], int n)
    {
        // Create a priority queue
        PriorityQueue<Integer> pq
            = new PriorityQueue<Integer>();

        // Adding items to the pQueue
        for (int i = 0; i < n; i++) {
            pq.add(arr[i]);
        }

        // Initialize result
        int res = 0;

        // While size of priority queue
        // is more than 1
        while (pq.size() > 1) {
            // Extract shortest two ropes from pq
            int first = pq.poll();
            int second = pq.poll();

            // Connect the ropes: update result
            // and insert the new rope to pq
            res += first + second;
            pq.add(first + second);
        }

        return res;
    }

    // Driver program to test above function
    public static void main(String args[])
    {
        int len[] = { 4, 3, 2, 6 };
        int size = len.length;
        System.out.println("Total cost for connecting"
                           + " ropes is "
                           + minCost(len, size));
    }
}
// This code is contributed by yash_pec

Python3

# Python3 program to connect n
# ropes with minimum cost
import heapq


def minCost(arr, n):

    # Create a priority queue out of the
    # given list
    heapq.heapify(arr)

    # Initialize result
    res = 0

    # While size of priority queue
    # is more than 1
    while(len(arr) > 1):

        # Extract shortest two ropes from arr
        first = heapq.heappop(arr)
        second = heapq.heappop(arr)

        # Connect the ropes: update result
        # and insert the new rope to arr
        res += first + second
        heapq.heappush(arr, first + second)

    return res


# Driver code
if __name__ == '__main__':

    lengths = [4, 3, 2, 6]
    size = len(lengths)

    print("Total cost for connecting ropes is " +
          str(minCost(lengths, size)))

# This code is contributed by shivampatel5

C#

// C# program to connect n
// ropes with minimum cost
using System;
using System.Collections.Generic;
public class ConnectRopes {
    static int minCost(int[] arr, int n)
    {
        // Create a priority queue
        List<int> pq = new List<int>();

        // Adding items to the pQueue
        for (int i = 0; i < n; i++) {
            pq.Add(arr[i]);
        }

        // Initialize result
        int res = 0;

        // While size of priority queue
        // is more than 1
        while (pq.Count > 1) {
            pq.Sort();

            // Extract shortest two ropes from pq
            int first = pq[0];
            int second = pq[1];
            pq.RemoveRange(0, 2);

            // Connect the ropes: update result
            // and insert the new rope to pq
            res += first + second;
            pq.Add(first + second);
        }
        return res;
    }

    // Driver program to test above function
    public static void Main(String[] args)
    {
        int[] len = { 4, 3, 2, 6 };
        int size = len.Length;
        Console.WriteLine("Total cost for connecting"
                          + " ropes is "
                          + minCost(len, size));
    }
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// JavaScript program to connect n
// ropes with minimum cost

function minCost(arr,n)
{
    // Create a priority queue
        let pq = [];

        // Adding items to the pQueue
        for (let i = 0; i < n; i++) {
            pq.push(arr[i]);
        }

        pq.sort(function(a,b){return a-b;});

        // Initialize result
        let res = 0;

        // While size of priority queue
        // is more than 1
        while (pq.length > 1) {
            // Extract shortest two ropes from pq
            let first = pq.shift();
            let second = pq.shift();

            // Connect the ropes: update result
            // and insert the new rope to pq
            res += first + second;
            pq.push(first + second);
            pq.sort(function(a,b){return a-b;});
        }

        return res;
}

// Driver program to test above function
let len = [4, 3, 2, 6];
let size = len.length;
document.write("Total cost for connecting"
                           + " ropes is " + minCost(len, size));

// This code is contributed by avanitrachhadiya2155

</script>

Output

Total cost for connecting ropes is 29

Time Complexity: O(N*log(N))
Auxiliary Space: O(N)