Introduction to Augmented Data Structure
Last Updated :
15 Feb, 2024
Data Structures play a significant role in building software and applications but many a times all our requirements are not satisfied using an existing data structure. This is when we modify an existing data structure according to our needs. This article will provide a brief introduction about when and how to Augment a Data Structure.
What is an Augmented Data Structure?
Augmenting a Data Structure (or Augmented Data Structure) means adapting an existing data structure to our needs. This allows us to take advantage of an original data structure that solves our problem partially and make changes such that it fits to our problem completely.
Examples of Augmenting a Data Structure:
There are many examples of augmenting data structures, and augmenting is simply changing the existing data structure to solve our problem.
- Augmented Queue for Maximum Element: In an augmented queue, each element not only stores its value but also maintains information about the maximum element in the current queue up to that point.
- Augmented Trie for Prefix Sums: Consider a trie (prefix tree) data structure where each node stores the sum of values of all elements with the same prefix up to that node.
- Augmented AVL Tree for Rank Queries: In an AVL tree (a self-balancing binary search tree), each node can store the rank of that node in the tree, i.e., the number of nodes in its left subtree plus one.
- Augmented Disjoint Set (Union-Find) for Set Size: In a disjoint set data structure, each set representative (root) could store the size of its corresponding set.
Prerequisites for Augmenting a Data Structure:
Augmenting a data structure involve adding new information or functionality to an existing data structure in order to improve its capabilities. While this can be a useful technique for improving an algorithm’s performance or functionality, it is critical to carefully consider the implications of augmenting a data structure before doing so. Here are some important considerations:
- Determine what additional information or functionality is required: Define precisely what information or functionality you want to add to the data structure. This could imply storing more data, keeping auxiliary information, or enabling new operations.
- Impact on performance and complexity: Augmentation may result in additional time and space complexity. Examine how the proposed augmentation will affect the operation of the data structure. Ensure that the benefits outweigh the performance costs.
- Consistency: The augmented data structure must be consistent with the original data structure. This means that the augmentation should not violate any of the original structure’s invariants or properties.
- Compatibility with existing code: If the data structure is already used in existing code, make sure that the augmentation does not break compatibility. Consider whether backward compatibility or migration strategies are required.
- Test and validate: The augmented data structure should be tested thoroughly to ensure its correctness, efficiency, and compatibility with existing code. Verify that the augmentation results in the desired performance or functionality improvements.
How to Augment a Data Structure?
The process of augmenting a data structure are as follows:
1. Define the Need:
- Identify the problem: Clearly define the problem you want the augmented structure to solve. What specific inefficiency or limitation are we addressing?
- Justify the augmentation: Is augmentation the most effective solution? Can simpler optimizations achieve the desired results?
2. Choose the Underlying Structure:
- Select the appropriate data structure: Consider the existing structure’s strengths and weaknesses. Can it efficiently support the augmentation you want to implement?
- Evaluate compatibility: Ensure the chosen structure allows efficient updates and maintenance of the added information during its basic operations (insertion, deletion, search, etc.).
3. Design the Augmentation:
- Define the data storage: Decide how the added information will be stored within the existing structure. Will it be part of each element, attached to nodes, or stored separately?
- Update operations: Modify existing operations (insertion, deletion, search, etc.) to incorporate the added information and ensure its consistency.
4. Implement and Test:
- Code the augmentation: Implement the designed modifications carefully, focusing on clarity, efficiency, and maintainability.
- Thorough testing: Test the augmented structure extensively for functionality, performance, and data integrity across various scenarios.
- Document and share: Document your design decisions, rationale, and limitations for future reference and to facilitate collaboration.
A Problem using Augmentation of Data Structure:
Given a set of integers, the task is to design a data structure that supports two operations efficiently:
- Insertion: Insert an integer into the data structure.
- Query: Given an integer
k
, find the number of elements in the data structure that are less than or equal to k
.
Naive Approach: A simple way to solve this problem is to use an array or a list to store the integers. For insertion, you can just append the integer to the end of the list. For the query operation, iterate through the list and count the elements less than or equal to k
. This approach takes linear time for the query operation.
Augmenting the Data Structure: To improve the efficiency of the query operation, we can augment the data structure with additional information. One way to do this is by maintaining a sorted order of elements. This can be achieved using a balanced Binary Search Tree (BST).
Structure of Augmented Tree: Each node in the BST should store the value of the element, the size of the subtree rooted at that node, and pointers to the left and right children.
- Insertion:
- Insert the element into the BST as you normally would.
- While inserting, update the size of each visited node.
- Query:
- Start at the root of the BST.
- Traverse the tree:
- If the current node’s value is less than or equal to
k
, add the size of its left subtree (including itself) to the count.
- Move to the left or right child accordingly.
Implementation of above algorithm is given below:
C++
#include <iostream>
class AugmentedBSTNode {
public :
int value;
int size;
AugmentedBSTNode* left;
AugmentedBSTNode* right;
AugmentedBSTNode( int value) {
this ->value = value;
this ->size = 1;
this ->left = nullptr;
this ->right = nullptr;
}
};
class AugmentedBST {
private :
AugmentedBSTNode* root;
AugmentedBSTNode* _insert(AugmentedBSTNode* node, int value) {
if (node == nullptr) {
return new AugmentedBSTNode(value);
}
node->size++;
if (value <= node->value) {
node->left = _insert(node->left, value);
} else {
node->right = _insert(node->right, value);
}
return node;
}
int _query(AugmentedBSTNode* node, int k) {
if (node == nullptr) {
return 0;
}
if (node->value <= k) {
return (node->left != nullptr ? node->left->size : 0) + 1 + _query(node->right, k);
} else {
return _query(node->left, k);
}
}
public :
AugmentedBST() {
this ->root = nullptr;
}
void insert( int value) {
root = _insert(root, value);
}
int query( int k) {
return _query(root, k);
}
};
int main() {
AugmentedBST augmentedDS;
augmentedDS.insert(5);
augmentedDS.insert(2);
augmentedDS.insert(8);
augmentedDS.insert(1);
int result = augmentedDS.query(6);
std::cout << result << std::endl;
return 0;
}
|
Java
class AugmentedBSTNode {
int value;
int size;
AugmentedBSTNode left;
AugmentedBSTNode right;
public AugmentedBSTNode( int value) {
this .value = value;
this .size = 1 ;
this .left = null ;
this .right = null ;
}
}
class AugmentedBST {
private AugmentedBSTNode root;
public void insert( int value) {
root = _insert(root, value);
}
private AugmentedBSTNode _insert(AugmentedBSTNode node, int value) {
if (node == null ) {
return new AugmentedBSTNode(value);
}
node.size++;
if (value <= node.value) {
node.left = _insert(node.left, value);
} else {
node.right = _insert(node.right, value);
}
return node;
}
public int query( int k) {
return _query(root, k);
}
private int _query(AugmentedBSTNode node, int k) {
if (node == null ) {
return 0 ;
}
if (node.value <= k) {
return (node.left != null ? node.left.size : 0 ) + 1 + _query(node.right, k);
} else {
return _query(node.left, k);
}
}
}
public class Main {
public static void main(String[] args) {
AugmentedBST augmentedDS = new AugmentedBST();
augmentedDS.insert( 5 );
augmentedDS.insert( 2 );
augmentedDS.insert( 8 );
augmentedDS.insert( 1 );
int result = augmentedDS.query( 6 );
System.out.println(result);
}
}
|
Python
class AugmentedBSTNode:
def __init__( self , value):
self .value = value
self .size = 1
self .left = None
self .right = None
class AugmentedBST:
def __init__( self ):
self .root = None
def insert( self , value):
self .root = self ._insert( self .root, value)
def _insert( self , node, value):
if not node:
return AugmentedBSTNode(value)
node.size + = 1
if value < = node.value:
node.left = self ._insert(node.left, value)
else :
node.right = self ._insert(node.right, value)
return node
def query( self , k):
return self ._query( self .root, k)
def _query( self , node, k):
if not node:
return 0
if node.value < = k:
return (node.left.size if node.left else 0 ) + 1 + self ._query(node.right, k)
else :
return self ._query(node.left, k)
augmented_ds = AugmentedBST()
augmented_ds.insert( 5 )
augmented_ds.insert( 2 )
augmented_ds.insert( 8 )
augmented_ds.insert( 1 )
result = augmented_ds.query( 6 )
print (result)
|
C#
using System;
class AugmentedBSTNode
{
public int value;
public int size;
public AugmentedBSTNode left;
public AugmentedBSTNode right;
public AugmentedBSTNode( int value)
{
this .value = value;
this .size = 1;
this .left = null ;
this .right = null ;
}
}
class AugmentedBST
{
private AugmentedBSTNode root;
private AugmentedBSTNode Insert(AugmentedBSTNode node, int value)
{
if (node == null )
{
return new AugmentedBSTNode(value);
}
node.size++;
if (value <= node.value)
{
node.left = Insert(node.left, value);
}
else
{
node.right = Insert(node.right, value);
}
return node;
}
private int Query(AugmentedBSTNode node, int k)
{
if (node == null )
{
return 0;
}
if (node.value <= k)
{
return (node.left != null ? node.left.size : 0) + 1 + Query(node.right, k);
}
else
{
return Query(node.left, k);
}
}
public AugmentedBST()
{
this .root = null ;
}
public void Insert( int value)
{
root = Insert(root, value);
}
public int Query( int k)
{
return Query(root, k);
}
}
class Program
{
static void Main()
{
AugmentedBST augmentedDS = new AugmentedBST();
augmentedDS.Insert(5);
augmentedDS.Insert(2);
augmentedDS.Insert(8);
augmentedDS.Insert(1);
int result = augmentedDS.Query(6);
Console.WriteLine(result);
}
}
|
Javascript
class AugmentedBSTNode {
constructor(value) {
this .value = value;
this .size = 1;
this .left = null ;
this .right = null ;
}
}
class AugmentedBST {
constructor() {
this .root = null ;
}
insert(value) {
this .root = this ._insert( this .root, value);
}
_insert(node, value) {
if (node === null ) {
return new AugmentedBSTNode(value);
}
node.size++;
if (value <= node.value) {
node.left = this ._insert(node.left, value);
} else {
node.right = this ._insert(node.right, value);
}
return node;
}
query(k) {
return this ._query( this .root, k);
}
_query(node, k) {
if (node === null ) {
return 0;
}
if (node.value <= k) {
return (node.left !== null ? node.left.size : 0) + 1 + this ._query(node.right, k);
} else {
return this ._query(node.left, k);
}
}
}
const augmentedDS = new AugmentedBST();
augmentedDS.insert(5);
augmentedDS.insert(2);
augmentedDS.insert(8);
augmentedDS.insert(1);
const result = augmentedDS.query(6);
console.log(result);
|
Time Complexity: O(Q * log(N)), where Q is the number of queries and N is the number of nodes in the BST. After Augmenting the BST, each query can be answered in O(logN) time.
Auxiliary Space: O(N)
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