Ropes Data Structure (Fast String Concatenation)
Last Updated :
17 Mar, 2023
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)
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Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
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