# Connect n ropes with minimum cost

There are given n ropes of different lengths, we need to connect these ropes into one rope. The cost to connect two ropes is equal to the sum of their lengths. We need to connect the ropes with minimum cost.

For example, if we are given 4 ropes of lengths 4, 3, 2, and 6. We can connect the ropes in the following ways.
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.

Total cost for connecting all ropes is 5 + 9 + 15 = 29. This is the optimized cost for connecting ropes. Other ways of connecting ropes would always have same or more cost. For example, if we connect 4 and 6 first (we get three strings of 3, 2 and 10), then connect 10 and 3 (we get two strings of 13 and 2). Finally we connect 13 and 2. Total cost in this way is 10 + 13 + 15 = 38.

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

Solution:
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 total cost. Therefore, the idea is to connect the smallest two ropes first and recur for remaining ropes. This approach is similar to Huffman Coding. We put the smallest ropes down the tree so that they can be repeated multiple times rather than the longer ropes.

So it forms a structure like a tree: The sum contains 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:

1. Create a min-heap and insert all lengths into the min-heap.
2. Do following while the number of elements in min-heap is not 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 this total cost.

Following is the implementation of the above algorithm.

## C++

 `// C++ program for connecting ` `// n ropes with minimum cost ` `#include ` ` `  `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; ``// Attay 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; ` `    ``minHeap->harr = 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); ` `    ``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; ` `        ``} ` ` `  `        ``// Store the minimum value, and remove it from heap ` `        ``int` `root = harr; ` `        ``harr = 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 `

Output:

```Total cost for connecting ropes is 29
```

Complexity Analysis:

• Time Complexity: O(nLogn), assuming that we use a O(nLogn) sorting algorithm. Note that heap operations like insert and extract take O(Logn) time.
• Auxiliary Complexity:O(n), The space required to store the values in min heap

A simple implementation with STL in C++
This uses priority_queue available in STL. Thanks to Pango89 for providing the below code. The approach and algorithm remains the same. The min Heap is replaced by a priority Queue.

## C++

 `#include ` ` `  `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); ` `    ``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 pq = ``new` `PriorityQueue(); ` ` `  `        ``// 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) ` `     `  `    ``# Initializ 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`

Output:

```Total cost for connecting ropes is 29
```

Complexity Analysis:

• Time Complexity: O(nLogn), assuming that we use a O(nLogn) sorting algorithm.
Note that heap operations like insert and extract take O(Logn) time.
• Auxiliary Complexity: O(n).
The space required to store the values in min heap