Ropes Data Structure (Fast String Concatenation)

One of the most common operations on strings is appending or concatenation. Appending to the end of a string when the string is stored in the traditional manner (i.e. an array of characters) would take a minimum of O(n) time (where n is the length of the original string). 
We can reduce time taken by append using Ropes Data Structure. 

Ropes Data Structure

A Rope is a binary tree structure where each node except the leaf nodes, contains the number of characters present to the left of that node. Leaf nodes contain the actual string broken into substrings (size of these substrings can be decided by the user).
Consider the image below.

The image shows how the string is stored in memory. Each leaf node contains substrings of the original string and all other nodes contain the number of characters present to the left of that node. The idea behind storing the number of characters to the left is to minimise the cost of finding the character present at i-th position.
Advantages
1. Ropes drastically cut down the cost of appending two strings. 
2. Unlike arrays, ropes do not require large contiguous memory allocations. 
3. Ropes do not require O(n) additional memory to perform operations like insertion/deletion/searching. 
4. In case a user wants to undo the last concatenation made, he can do so in O(1) time by just removing the root node of the tree.
Disadvantages
1. The complexity of source code increases. 
2. Greater chances of bugs. 
3. Extra memory required to store parent nodes. 
4. Time to access i-th character increases.
Now let’s look at a situation that explains why Ropes are a good substitute to monolithic string arrays.
Given two strings a[] and b[]. Concatenate them in a third string c[].
Examples: 

Input  : a[] = "This is ", b[] = "an apple"
Output : "This is an apple"

Input  : a[] = "This is ", b[] = "geeksforgeeks"
Output : "This is geeksforgeeks"

Method 1 (Naive method)

We create a string c[] to store concatenated string. We first traverse a[] and copy all characters of a[] to c[]. Then we copy all characters of b[] to c[]. 
Implementation:

C++

// Simple C++ program to concatenate two strings
#include <iostream>
using namespace std;

// Function that concatenates strings a[0..n1-1]
// and b[0..n2-1] and stores the result in c[]
void concatenate(char a[], char b[], char c[],
                              int n1, int n2)
{
    // Copy characters of A[] to C[]
    int i;
    for (i=0; i<n1; i++)
        c[i] = a[i];

    // Copy characters of B[]
    for (int j=0; j<n2; j++)
        c[i++] = b[j];

    c[i] = '\0';
}


// Driver code
int main()
{
    char a[] =  "Hi This is geeksforgeeks. ";
    int n1 = sizeof(a)/sizeof(a[0]);

    char b[] =  "You are welcome here.";
    int n2 = sizeof(b)/sizeof(b[0]);

    // Concatenate a[] and b[] and store result
    // in c[]
    char c[n1 + n2 - 1];
    concatenate(a, b, c, n1, n2);
    for (int i=0; i<n1+n2-1; i++)
        cout << c[i];

    return 0;
}

Java

//Java program to concatenate two strings

class GFG {

    // Function that concatenates strings a[0..n1-1]
    // and b[0..n2-1] and stores the result in c[]
    static void concatenate(char a[], char b[], char c[],
            int n1, int n2) {
        // Copy characters of A[] to C[]
        int i;
        for (i = 0; i < n1; i++) {
            c[i] = a[i];
        }

        // Copy characters of B[]
        for (int j = 0; j < n2; j++) {
            c[i++] = b[j];
        }

    }

    // Driver code
    public static void main(String[] args) {
        char a[] = "Hi This is geeksforgeeks. ".toCharArray();
        int n1 = a.length;

        char b[] = "You are welcome here.".toCharArray();
        int n2 = b.length;

        // Concatenate a[] and b[] and store result
        // in c[]
        char c[] = new char[n1 + n2];
        concatenate(a, b, c, n1, n2);
        for (int i = 0; i < n1 + n2 - 1; i++) {
            System.out.print(c[i]);
        }

    }
}
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to concatenate two strings

# Function that concatenates strings a[0..n1-1]
# and b[0..n2-1] and stores the result in c[]
def concatenate(a, b, c, n1, n2):

    # Copy characters of A[] to C[]
    i = -1
    for i in range(n1):
        c[i] = a[i]

    # Copy characters of B[]
    for j in range(n2):
        c[i] = b[j]
        i += 1

# Driver Code
if __name__ == "__main__":
    a = "Hi This is geeksforgeeks. "
    n1 = len(a)

    b = "You are welcome here."
    n2 = len(b)

    a = list(a)
    b = list(b)

    # Concatenate a[] and b[] and
    # store result in c[]
    c = [0] * (n1 + n2 - 1)

    concatenate(a, b, c, n1, n2)

    for i in c:
        print(i, end = "")

# This code is contributed by
# sanjeev2552

C#

// C# program to concatenate two strings
using System;

public class GFG {

    // Function that concatenates strings a[0..n1-1]
    // and b[0..n2-1] and stores the result in c[]
    static void concatenate(char []a, char []b, char []c,
            int n1, int n2) {
        // Copy characters of A[] to C[]
        int i;
        for (i = 0; i < n1; i++) {
            c[i] = a[i];
        }

        // Copy characters of B[]
        for (int j = 0; j < n2; j++) {
            c[i++] = b[j];
        }

    }

    // Driver code
    public static void Main() {
        char []a = "Hi This is geeksforgeeks. ".ToCharArray();
        int n1 = a.Length;

        char []b = "You are welcome here.".ToCharArray();
        int n2 = b.Length;

        // Concatenate a[] and b[] and store result
        // in c[]
        char []c = new char[n1 + n2];
        concatenate(a, b, c, n1, n2);
        for (int i = 0; i < n1 + n2 - 1; i++) {
            Console.Write(c[i]);
        }

    }
}
/*This code is contributed by PrinciRaj1992*/

Javascript

// Function that concatenates strings a[0..n1-1]
// and b[0..n2-1] and stores the result in c[]
function concatenate(a, b, c, n1, n2) {
    // Copy characters of A[] to C[]
    let i;
    for (i=0; i<n1; i++)
        c[i] = a[i];

    // Copy characters of B[]
    for (let j=0; j<n2; j++)
        c[i++] = b[j];

    c[i] = '\0';
}


// Driver code
function main() {
    let a =  "Hi This is geeksforgeeks. ";
    let n1 = a.length;

    let b =  "You are welcome here.";
    let n2 = b.length;

    // Concatenate a[] and b[] and store result
    // in c[]
    let c = Array(n1 + n2 - 1);
    concatenate(a, b, c, n1, n2);
    for (let i=0; i<n1+n2-1; i++)
        console.log(c[i]);

    return 0;
}
main();

Output:

Hi This is geeksforgeeks. You are welcome here

Time complexity : O(max(n1, n2))
Auxiliary Space: O(n1 + n2)

Now let’s try to solve the same problem using Ropes.

Method 2 (Rope structure method)

This rope structure can be utilized to concatenate two strings in constant time. 
1. Create a new root node (that stores the root of the new concatenated string) 
2. Mark the left child of this node, the root of the string that appears first. 
3. Mark the right child of this node, the root of the string that appears second.
And that’s it. Since this method only requires to make a new node, it’s complexity is O(1).
Consider the image below (Image source : https://en.wikipedia.org/wiki/Rope_(data_structure))

Implementation:

CPP

// C++ program to concatenate two strings using
// rope data structure.
#include <bits/stdc++.h>
using namespace std;

// Maximum no. of characters to be put in leaf nodes
const int LEAF_LEN = 2;

// Rope structure
class Rope
{
public:
    Rope *left, *right, *parent;
    char *str;
    int lCount;
};

// Function that creates a Rope structure.
// node --> Reference to pointer of current root node
//   l  --> Left index of current substring (initially 0)
//   r  --> Right index of current substring (initially n-1)
//   par --> Parent of current node (Initially NULL)
void createRopeStructure(Rope *&node, Rope *par,
                         char a[], int l, int r)
{
    Rope *tmp = new Rope();
    tmp->left = tmp->right = NULL;

    // We put half nodes in left subtree
    tmp->parent = par;

    // If string length is more
    if ((r-l) > LEAF_LEN)
    {
        tmp->str = NULL;
        tmp->lCount = (r-l)/2;
        node = tmp;
        int m = (l + r)/2;
        createRopeStructure(node->left, node, a, l, m);
        createRopeStructure(node->right, node, a, m+1, r);
    }
    else
    {
        node = tmp;
        tmp->lCount = (r-l);
        int j = 0;
        tmp->str = new char[LEAF_LEN];
        for (int i=l; i<=r; i++)
            tmp->str[j++] = a[i];
    }
}

// Function that prints the string (leaf nodes)
void printstring(Rope *r)
{
    if (r==NULL)
        return;
    if (r->left==NULL && r->right==NULL)
        cout << r->str;
    printstring(r->left);
    printstring(r->right);
}

// Function that efficiently concatenates two strings
// with roots root1 and root2 respectively. n1 is size of
// string represented by root1.
// root3 is going to store root of concatenated Rope.
void concatenate(Rope *&root3, Rope *root1, Rope *root2, int n1)
{
    // Create a new Rope node, and make root1
    // and root2 as children of tmp.
    Rope *tmp = new Rope();
    tmp->parent = NULL;
    tmp->left = root1;
    tmp->right = root2;
    root1->parent = root2->parent = tmp;
    tmp->lCount = n1;

    // Make string of tmp empty and update
    // reference r
    tmp->str = NULL;
    root3 = tmp;
}

// Driver code
int main()
{
    // Create a Rope tree for first string
    Rope *root1 = NULL;
    char a[] =  "Hi This is geeksforgeeks. ";
    int n1 = sizeof(a)/sizeof(a[0]);
    createRopeStructure(root1, NULL, a, 0, n1-1);

    // Create a Rope tree for second string
    Rope *root2 = NULL;
    char b[] =  "You are welcome here.";
    int n2 = sizeof(b)/sizeof(b[0]);
    createRopeStructure(root2, NULL, b, 0, n2-1);

    // Concatenate the two strings in root3.
    Rope *root3 = NULL;
    concatenate(root3, root1, root2, n1);

    // Print the new concatenated string
    printstring(root3);
    cout << endl;
    return 0;
}

Java

import java.util.ArrayList;
// Rope structure
class Rope {
    Rope left;
    Rope right;
    Rope parent;
    ArrayList<Character> str;
    int lCount;

    Rope()
    {
        this.left = null;
        this.right = null;
        this.parent = null;
        this.str = new ArrayList<Character>();
        this.lCount = 0;
    }
}

class Main {
    // Maximum no. of characters to be put in leaf nodes
    static final int LEAF_LEN = 2;
    // Function that creates a Rope structure.
    // node --> Reference to pointer of current root node
    //   l  --> Left index of current substring (initially
    //   0) r  --> Right index of current substring
    //   (initially n-1) par --> Parent of current node
    //   (Initially NULL)
    static Rope createRopeStructure(Rope node, Rope par,
                                    String a, int l, int r)
    {
        Rope tmp = new Rope();
        tmp.left = tmp.right = null;
        // We put half nodes in left subtree
        tmp.parent = par;

        if ((r - l) > LEAF_LEN) {
            tmp.str = null;
            tmp.lCount = (int)Math.floor((r - l) / 2);
            node = tmp;
            int m = (int)Math.floor((l + r) / 2);
            node.left = createRopeStructure(node.left, node,
                                            a, l, m);
            node.right = createRopeStructure(
                node.right, node, a, m + 1, r);
        }
        else {
            node = tmp;
            tmp.lCount = (r - l);
            int j = 0;
            for (int i = l; i <= r; i++) {
                tmp.str.add(a.charAt(i));
            }
        }

        return node;
    }
    // Function that prints the string (leaf nodes)
    static void printstring(Rope r)
    {
        if (r == null) {
            return;
        }
        if (r.left == null && r.right == null) {
            for (char c : r.str) {
                System.out.print(c);
            }
        }
        printstring(r.left);
        printstring(r.right);
    }
    // Function that efficiently concatenates two strings
    // with roots root1 and root2 respectively. n1 is size
    // of string represented by root1. root3 is going to
    // store root of concatenated Rope.
    static Rope concatenate(Rope root3, Rope root1,
                            Rope root2, int n1)
    {
        // Create a new Rope node, and make root1
        // and root2 as children of tmp.
        Rope tmp = new Rope();
        tmp.left = root1;
        tmp.right = root2;
        root1.parent = tmp;
        root2.parent = tmp;
        tmp.lCount = n1;
        // Make string of tmp empty and update
        // reference r
        tmp.str = null;
        root3 = tmp;

        return root3;
    }
    // Driver code
    public static void main(String[] args)
    {
        // Create a Rope tree for first string
        Rope root1 = null;
        String a = "Hi This is geeksforgeeks. ";
        int n1 = a.length();
        root1 = createRopeStructure(root1, null, a, 0,
                                    n1 - 1);
        // Create a Rope tree for second string
        Rope root2 = null;
        String b = "You are welcome here.";
        int n2 = b.length();
        root2 = createRopeStructure(root2, null, b, 0,
                                    n2 - 1);

        // Concatenate the two strings in root3.
        Rope root3 = null;
        root3 = concatenate(root3, root1, root2, n1);
        // Print the new concatenated string
        printstring(root3);
    }
}

Python3

# Python program to concatenate two strings using
# rope data structure.

# Maximum no. of characters to be put in leaf nodes
LEAF_LEN = 2

# Rope structure
class Rope:

    def __init__(self):
        self.left = None
        self.right = None
        self.parent = None
        self.str = [0]*(LEAF_LEN + 1)
        self.lCount = 0

# Function that creates a Rope structure.
# node --> Reference to pointer of current root node
# l  --> Left index of current substring (initially 0)
# r  --> Right index of current substring (initially n-1)
# par --> Parent of current node (Initially NULL)
def createRopeStructure(node, par, a, l, r):

    tmp = Rope()
    tmp.left = tmp.right = None

    # We put half nodes in left subtree
    tmp.parent = par

    # If string length is more
    if (r-l) > LEAF_LEN:

        tmp.str = None
        tmp.lCount = (r-l) // 2
        node = tmp
        m = (l + r) // 2
        createRopeStructure(node.left, node, a, l, m)
        createRopeStructure(node.right, node, a, m+1, r)
    else:

        node = tmp
        tmp.lCount = (r-l)
        j = 0
        for i in range(l, r+1):
            print(a[i],end = "")
            tmp.str[j] = a[i]
            j = j + 1
        print(end = "")

    return node

# Function that prints the string (leaf nodes)
def printstring(r):

    if r==None:
        return

    if r.left==None and r.right==None:
        # console.log(r.str);
        pass

    printstring(r.left)
    printstring(r.right)

# Function that efficiently concatenates two strings
# with roots root1 and root2 respectively. n1 is size of
# string represented by root1.
# root3 is going to store root of concatenated Rope.
def concatenate(root3, root1, root2, n1):

    # Create a new Rope node, and make root1
    # and root2 as children of tmp.
    tmp = Rope()
    tmp.left = root1
    tmp.right = root2
    root1.parent = tmp
    root2.parent = tmp
    tmp.lCount = n1

    # Make string of tmp empty and update
    # reference r
    tmp.str = None
    root3 = tmp

    return root3

# Driver code
# Create a Rope tree for first string
root1 = None
a =  "Hi This is geeksforgeeks. "
n1 = len(a)
root1 = createRopeStructure(root1, None, a, 0, n1-1)

# Create a Rope tree for second string
root2 = None
b =  "You are welcome here."
n2 = len(b)
root2 = createRopeStructure(root2, None, b, 0, n2-1)

# Concatenate the two strings in root3.
root3 = None
root3 = concatenate(root3, root1, root2, n1)

# Print the new concatenated string
printstring(root3)
print()

# The code is contributed by Nidhi goel.

Javascript

// javascript program to concatenate two strings using
// rope data structure.

// Maximum no. of characters to be put in leaf nodes
const LEAF_LEN = 2;

// Rope structure
class Rope
{
    constructor(){
        this.left = null;
        this.right = null;
        this.parent = null;
        this.str = new Array();
        this.lCount = 0;
    }
}

// Function that creates a Rope structure.
// node --> Reference to pointer of current root node
//   l  --> Left index of current substring (initially 0)
//   r  --> Right index of current substring (initially n-1)
//   par --> Parent of current node (Initially NULL)
function createRopeStructure(node, par, a, l, r)
{
    let tmp = new Rope();
    tmp.left = tmp.right = null;

    // We put half nodes in left subtree
    tmp.parent = par;

    // If string length is more
    if ((r-l) > LEAF_LEN)
    {
        tmp.str = null;
        tmp.lCount = Math.floor((r-l)/2);
        node = tmp;
        let m = Math.floor((l + r)/2);
        createRopeStructure(node.left, node, a, l, m);
        createRopeStructure(node.right, node, a, m+1, r);
    }
    else
    {
        node = tmp;
        tmp.lCount = (r-l);
        let j = 0;
        // tmp.str = new Array(LEAF_LEN);
        for (let i=l; i<=r; i++){
            document.write(a[i]);
            tmp.str[j++] = a[i];
        }
        document.write("\n");
    }

    return node;
}

// Function that prints the string (leaf nodes)
function printstring(r)
{
    if (r==null)
        return;
    if (r.left==null && r.right==null){
        // console.log(r.str);
    }

    printstring(r.left);
    printstring(r.right);
}

// Function that efficiently concatenates two strings
// with roots root1 and root2 respectively. n1 is size of
// string represented by root1.
// root3 is going to store root of concatenated Rope.
function concatenate(root3, root1, root2, n1)
{
    // Create a new Rope node, and make root1
    // and root2 as children of tmp.
    let tmp = new Rope();
    tmp.left = root1;
    tmp.right = root2;
    root1.parent = tmp;
    root2.parent = tmp;
    tmp.lCount = n1;

    // Make string of tmp empty and update
    // reference r
    tmp.str = null;
    root3 = tmp;

    return root3;
}

// Driver code
// Create a Rope tree for first string
let root1 = null;
let a =  "Hi This is geeksforgeeks. ";
let n1 = a.length;
root1 = createRopeStructure(root1, null, a, 0, n1-1);

// Create a Rope tree for second string
let root2 = null;
let b =  "You are welcome here.";
let n2 = b.length;
root2 = createRopeStructure(root2, null, b, 0, n2-1);

// Concatenate the two strings in root3.
let root3 = null;
root3 = concatenate(root3, root1, root2, n1);

// Print the new concatenated string
printstring(root3);
console.log();

// The code is contributed by Nidhi goel.

C#

using System;
using System.Collections.Generic;

// Rope structure
class Rope
{
    public Rope left;
    public Rope right;
    public Rope parent;
    public List<char> str;
    public int lCount;

    public Rope()
    {
        this.left = null;
        this.right = null;
        this.parent = null;
        this.str = new List<char>();
        this.lCount = 0;
    }
}

class MainClass
{
    // Maximum no. of characters to be put in leaf nodes
    static readonly int LEAF_LEN = 2;

    // Function that creates a Rope structure.
    // node --> Reference to pointer of current root node
    //   l  --> Left index of current substring (initially
    //   0) r  --> Right index of current substring
    //   (initially n-1) par --> Parent of current node
    //   (Initially NULL)
    static Rope CreateRopeStructure(ref Rope node, Rope par, string a, int l, int r)
    {
        Rope tmp = new Rope();
        tmp.left = tmp.right = null;
        // We put half nodes in left subtree
        tmp.parent = par;

        if ((r - l) > LEAF_LEN)
        {
            tmp.str = null;
            tmp.lCount = (int)Math.Floor((r - l) / 2.0);
            node = tmp;
            int m = (int)Math.Floor((l + r) / 2.0);
            node.left = CreateRopeStructure(ref node.left, node, a, l, m);
            node.right = CreateRopeStructure(ref node.right, node, a, m + 1, r);
        }
        else
        {
            node = tmp;
            tmp.lCount = (r - l);
            for (int i = l; i <= r; i++)
            {
                tmp.str.Add(a[i]);
            }
        }

        return node;
    }

    // Function that prints the string (leaf nodes)
    static void PrintString(Rope r)
    {
        if (r == null)
        {
            return;
        }
        if (r.left == null && r.right == null)
        {
            foreach (char c in r.str)
            {
                Console.Write(c);
            }
        }
        PrintString(r.left);
        PrintString(r.right);
    }

    // Function that efficiently concatenates two strings
    // with roots root1 and root2 respectively. n1 is size
    // of string represented by root1. root3 is going to
    // store root of concatenated Rope.
    static Rope Concatenate(Rope root3, Rope root1, Rope root2, int n1)
    {
        // Create a new Rope node, and make root1
        // and root2 as children of tmp.
        Rope tmp = new Rope();
        tmp.left = root1;
        tmp.right = root2;
        root1.parent = tmp;
        root2.parent = tmp;
        tmp.lCount = n1;
        // Make string of tmp empty and update
        // reference r
        tmp.str = null;
        root3 = tmp;

        return root3;
    }

    // Driver code
    public static void Main(string[] args)
    {
        // Create a Rope tree for first string
        Rope root1 = null;
        string a = "Hi This is geeksforgeeks. ";
        int n1 = a.Length;
        root1 = CreateRopeStructure(ref root1, null, a, 0, n1 - 1);
        // Create a Rope tree for second string
        Rope root2 = null;
        String b = "You are welcome here.";
        int n2 = b.Length;
        root2 = CreateRopeStructure(ref root2, null, b, 0,
                                    n2 - 1);
         // Concatenate the two strings in root3.
        Rope root3 = null;
        root3 = Concatenate(root3, root1, root2, n1);
        // Print the new concatenated string
        PrintString(root3);

    }
}

Output:

Hi This is geeksforgeeks. You are welcome here.

Time Complexity: O(1) 
Auxiliary Space: O(1)

This article is contributed by Akhil Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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