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.
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[] = "w3wiki" Output : "This is w3wiki"
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 w3wiki. " ; 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 w3wiki. " .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 w3wiki. " 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 w3wiki. " .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 w3wiki. " ; 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 w3wiki. 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.
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 w3wiki. " ; 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 w3wiki. " ; 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 w3wiki. " 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 w3wiki. " ; 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 w3wiki. " ; 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 w3wiki. You are welcome here.
Time Complexity: O(1)
Auxiliary Space: O(1)
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