How to do the following operations efficiently if there are large number of queries for them.
- Insertion
- Deletion
- Searching
- Clearing/Removing all the elements.
One solution is to use a Self-Balancing Binary Search Tree like Red-Black Tree, AVL Tree, etc. Time complexity of this solution for insertion, deletion and searching is O(Log n).
We can also use Hashing. With hashing, time complexity of first three operations is O(1). But time complexity of the fourth operation is O(n).
We can also use bit-vector (or direct access table), but bit-vector also requires O(n) time for clearing.
Sparse Set outperforms all BST, Hashing and bit vector. We assume that we are given range of data (or maximum value an element can have) and maximum number of elements that can be stored in set. The idea is to maintain two arrays: sparse[] and dense[].
dense[] ==> Stores the actual elements sparse[] ==> This is like bit-vector where we use elements as index. Here values are not binary, but indexes of dense array. maxVal ==> Maximum value this set can store. Size of sparse[] is equal to maxVal + 1. capacity ==> Capacity of Set. Size of sparse is equal to capacity. n ==> Current number of elements in Set.
insert(x): Let x be the element to be inserted. If x is greater than maxVal or n (current number of elements) is greater than equal to capacity, we return.
If none of the above conditions is true, we insert x in dense[] at index n (position after last element in a 0 based indexed array), increment n by one (Current number of elements) and store n (index of x in dense[]) at sparse[x].
search(x): To search an element x, we use x as index in sparse[]. The value sparse[x] is used as index in dense[]. And if value of dense[sparse[x]] is equal to x, we return dense[x]. Else we return -1.
delete(x): To delete an element x, we replace it with last element in dense[] and update index of last element in sparse[]. Finally decrement n by 1.
clear(): Set n = 0.
print(): We can print all elements by simply traversing dense[].
Illustration:
Let there be a set with two elements {3, 5}, maximum value as 10 and capacity as 4. The set would be represented as below. Initially: maxVal = 10 // Size of sparse capacity = 4 // Size of dense n = 2 // Current number of elements in set // dense[] Stores actual elements dense[] = {3, 5, _, _} // Uses actual elements as index and stores // indexes of dense[] sparse[] = {_, _, _, 0, _, 1, _, _, _, _,} '_' means it can be any value and not used in sparse set Insert 7: n = 3 dense[] = {3, 5, 7, _} sparse[] = {_, _, _, 0, _, 1, _, 2, _, _,} Insert 4: n = 4 dense[] = {3, 5, 7, 4} sparse[] = {_, _, _, 0, 3, 1, _, 2, _, _,} Delete 3: n = 3 dense[] = {4, 5, 7, _} sparse[] = {_, _, _, _, 0, 1, _, 2, _, _,} Clear (Remove All): n = 0 dense[] = {_, _, _, _} sparse[] = {_, _, _, _, _, _, _, _, _, _,}
Below is the implementation of above functions.
/* A C++ program to implement Sparse Set and its operations */ #include<bits/stdc++.h> using namespace std;
// A structure to hold the three parameters required to // represent a sparse set. class SSet
{ int *sparse; // To store indexes of actual elements
int *dense; // To store actual set elements
int n; // Current number of elements
int capacity; // Capacity of set or size of dense[]
int maxValue; /* Maximum value in set or size of
sparse[] */
public :
// Constructor
SSet( int maxV, int cap)
{
sparse = new int [maxV+1];
dense = new int [cap];
capacity = cap;
maxValue = maxV;
n = 0; // No elements initially
}
// Destructor
~SSet()
{
delete [] sparse;
delete [] dense;
}
// If element is present, returns index of
// element in dense[]. Else returns -1.
int search( int x);
// Inserts a new element into set
void insert( int x);
// Deletes an element
void deletion( int x);
// Prints contents of set
void print();
// Removes all elements from set
void clear() { n = 0; }
// Finds intersection of this set with s
// and returns pointer to result.
SSet* intersection(SSet &s);
// A function to find union of two sets
// Time Complexity-O(n1+n2)
SSet *setUnion(SSet &s);
}; // If x is present in set, then returns index // of it in dense[], else returns -1. int SSet::search( int x)
{ // Searched element must be in range
if (x > maxValue)
return -1;
// The first condition verifies that 'x' is
// within 'n' in this set and the second
// condition tells us that it is present in
// the data structure.
if (sparse[x] < n && dense[sparse[x]] == x)
return (sparse[x]);
// Not found
return -1;
} // Inserts a new element into set void SSet::insert( int x)
{ // Corner cases, x must not be out of
// range, dense[] should not be full and
// x should not already be present
if (x > maxValue)
return ;
if (n >= capacity)
return ;
if (search(x) != -1)
return ;
// Inserting into array-dense[] at index 'n'.
dense[n] = x;
// Mapping it to sparse[] array.
sparse[x] = n;
// Increment count of elements in set
n++;
} // A function that deletes 'x' if present in this data // structure, else it does nothing (just returns). // By deleting 'x', we unset 'x' from this set. void SSet::deletion( int x)
{ // If x is not present
if (search(x) == -1)
return ;
int temp = dense[n-1]; // Take an element from end
dense[sparse[x]] = temp; // Overwrite.
sparse[temp] = sparse[x]; // Overwrite.
// Since one element has been deleted, we
// decrement 'n' by 1.
n--;
} // prints contents of set which are also content // of dense[] void SSet::print()
{ for ( int i=0; i<n; i++)
printf ( "%d " , dense[i]);
printf ( "\n" );
} // A function to find intersection of two sets SSet* SSet::intersection(SSet &s) { // Capacity and max value of result set
int iCap = min(n, s.n);
int iMaxVal = max(s.maxValue, maxValue);
// Create result set
SSet *result = new SSet(iMaxVal, iCap);
// Find the smaller of two sets
// If this set is smaller
if (n < s.n)
{
// Search every element of this set in 's'.
// If found, add it to result
for ( int i = 0; i < n; i++)
if (s.search(dense[i]) != -1)
result->insert(dense[i]);
}
else
{
// Search every element of 's' in this set.
// If found, add it to result
for ( int i = 0; i < s.n; i++)
if (search(s.dense[i]) != -1)
result->insert(s.dense[i]);
}
return result;
} // A function to find union of two sets // Time Complexity-O(n1+n2) SSet* SSet::setUnion(SSet &s) { // Find capacity and maximum value for result
// set.
int uCap = s.n + n;
int uMaxVal = max(s.maxValue, maxValue);
// Create result set
SSet *result = new SSet(uMaxVal, uCap);
// Traverse the first set and insert all
// elements of it in result.
for ( int i = 0; i < n; i++)
result->insert(dense[i]);
// Traverse the second set and insert all
// elements of it in result (Note that sparse
// set doesn't insert an entry if it is already
// present)
for ( int i = 0; i < s.n; i++)
result->insert(s.dense[i]);
return result;
} // Driver program int main()
{ // Create a set set1 with capacity 5 and max
// value 100
SSet s1(100, 5);
// Insert elements into the set set1
s1.insert(5);
s1.insert(3);
s1.insert(9);
s1.insert(10);
// Printing the elements in the data structure.
printf ( "The elements in set1 are\n" );
s1.print();
int index = s1.search(3);
// 'index' variable stores the index of the number to
// be searched.
if (index != -1) // 3 exists
printf ( "\n3 is found at index %d in set1\n" ,index);
else // 3 doesn't exist
printf ( "\n3 doesn't exists in set1\n" );
// Delete 9 and print set1
s1.deletion(9);
s1.print();
// Create a set with capacity 6 and max value
// 1000
SSet s2(1000, 6);
// Insert elements into the set
s2.insert(4);
s2.insert(3);
s2.insert(7);
s2.insert(200);
// Printing set 2.
printf ( "\nThe elements in set2 are\n" );
s2.print();
// Printing the intersection of the two sets
SSet *intersect = s2.intersection(s1);
printf ( "\nIntersection of set1 and set2\n" );
intersect->print();
// Printing the union of the two sets
SSet *unionset = s1.setUnion(s2);
printf ( "\nUnion of set1 and set2\n" );
unionset->print();
return 0;
} |
// SSet class represents the Sparse Set data structure class SSet {
private int [] sparse; // array to store the index of elements in dense array
private int [] dense; // array to store the actual elements
private int capacity; // maximum capacity of the set
private int maxValue; // maximum value that can be stored in the set
private int n; // number of elements in the set
// Constructor to create an SSet object with given max value and capacity
public SSet( int maxV, int cap) {
sparse = new int [maxV + 1 ]; // create a sparse array with max value + 1
dense = new int [cap]; // create a dense array with the given capacity
capacity = cap;
maxValue = maxV;
n = 0 ; // initially the set is empty
}
// Search for an element in the set and return its index
public int search( int x) {
// check if the given element is out of range
if (x > maxValue) {
return - 1 ; // if yes, return -1
}
// check if the element exists in the set
if (sparse[x] < n && dense[sparse[x]] == x) {
return sparse[x]; // if yes, return its index
}
return - 1 ; // if not, return -1
}
// Insert an element into the set
public void insert( int x) {
// check if the element is out of range or the set is full or
// the element already exists in the set
if (x > maxValue || n >= capacity || search(x) != - 1 ) {
return ; // if yes, do nothing and return
}
// add the element to the end of the dense array
dense[n] = x;
// update the index of the element in the sparse array
sparse[x] = n;
n++; // increment the size of the set
}
// Delete an element from the set
public void deletion( int x) {
int index = search(x); // find the index of the element
// check if the element exists in the set
if (index == - 1 ) {
return ; // if not, do nothing and return
}
// swap the element with the last element in the dense array
int temp = dense[n - 1 ];
dense[index] = temp;
sparse[temp] = index;
n--; // decrement the size of the set
}
// Print the elements in the set
public void printSet() {
// print the elements in the dense array
for ( int i = 0 ; i < n; i++) {
System.out.print(dense[i] + " " ); // print the elements in the dense array
}
System.out.println();
}
public SSet intersection(SSet s) {
// Calculate the maximum size and capacity of the intersection set
int iCap = Math.min(n, s.n);
int iMaxVal = Math.max(maxValue, s.maxValue);
// Create the intersection set with the calculated size and capacity
SSet result = new SSet(iMaxVal, iCap);
// Check if the current set has fewer elements than the other set
if (n < s.n) {
// Iterate over each element in the current set
for ( int i = 0 ; i < n; i++) {
// Check if the current element is also in the other set
if (s.search(dense[i]) != - 1 ) {
result.insert(dense[i]);
}
}
} else {
// Iterate over each element in the other set
for ( int i = 0 ; i < s.n; i++) {
// Check if the current element is also in the current set
if (search(s.dense[i]) != - 1 ) {
result.insert(s.dense[i]);
}
}
}
// Return the intersection set
return result;
}
public SSet setUnion(SSet set2) {
// Calculate the maximum size and capacity of the union set
int uCap = n + set2.n;
int uMaxVal = Math.max(maxValue, set2.maxValue);
// Create the union set with the calculated size and capacity
SSet unionSet = new SSet(uMaxVal, uCap);
// Insert all elements from the current set into the union set
for ( int i = 0 ; i < n; i++) {
unionSet.insert(dense[i]);
}
// Insert all elements from the other set into the union set
for ( int i = 0 ; i < set2.n; i++) {
unionSet.insert(set2.dense[i]);
}
// Return the union set
return unionSet;
}
} public class Main {
public static void main(String[] args) {
// Create a set set1 with capacity 5 and max value 100
SSet s1 = new SSet( 100 , 5 );
// Insert elements into the set set1
s1.insert( 5 );
s1.insert( 3 );
s1.insert( 9 );
s1.insert( 10 );
// Printing the elements in the data structure
System.out.println( "The elements in set1 are:" );
s1.printSet();
int index = s1.search( 3 );
// 'index' variable stores the index of the number to be searched.
if (index != - 1 ) { // 3 exists
System.out.printf( "\n3 is found at index %d in set1" , index);
} else { // 3 doesn't exist
System.out.println( "\n3 doesn't exist in set1" );
}
// Delete 9 and print set1
s1.deletion( 9 );
s1.printSet();
// Create a set with capacity 6 and max value 1000
SSet s2 = new SSet( 1000 , 6 );
// Insert elements into the set
s2.insert( 4 );
s2.insert( 3 );
s2.insert( 7 );
s2.insert( 200 );
// Printing set 2
System.out.println( "\nThe elements in set2 are:" );
s2.printSet();
// Printing the intersection of the two sets
SSet intersect = s2.intersection(s1);
System.out.println( "\nIntersection of set1 and set2" );
intersect.printSet();
// Printing the union of the two sets
SSet unionset = s1.setUnion(s2);
System.out.println( "\nUnion of set1 and set2" );
unionset.printSet();
}
} // This code is contributed by amit_mangal_ |
# A python program to implement Sparse set and its operations. class SSet:
def __init__( self , maxV, cap):
self .sparse = [ 0 ] * (maxV + 1 ) # To store indexes of actual elements.
self .dense = [ 0 ] * cap # To store actual sets elements.
self .capacity = cap # capacity of set or size of dense
self .maxValue = maxV # Maximum value in set or size of sparse
self .n = 0 # Current no of elements, No elements initially
# If element is present, returns index of
# element in dense[]. Else returns -1.
def search( self , x):
# Searched element must be in range
if x > self .maxValue:
return - 1
# The first condition verifies that 'x' is
# within 'n' in this set and the second
# condition tells us that it is present in
# the data structure.
if self .sparse[x] < self .n and self .dense[ self .sparse[x]] = = x:
return self .sparse[x]
# Not found
return - 1
# Inserts a new element into set
def insert( self , x):
# Corner cases, x must not be out of
# range, dense[] should not be full and
# x should not already be present
if x > self .maxValue:
return
if self .n > = self .capacity:
return
if self .search(x) ! = - 1 :
return
# Inserting into array-dense[] at index 'n'.
self .dense[ self .n] = x
# Mapping it to sparse[] array.
self .sparse[x] = self .n
# Increment count of elements in set
self .n + = 1
# A function that deletes 'x' if present in this data
# structure, else it does nothing (just returns).
# By deleting 'x', we unset 'x' from this set.
def deletion( self , x):
# If x is not present
if self .search(x) = = - 1 :
return
temp = self .dense[ self .n - 1 ] # Take an element from end
self .dense[ self .sparse[x]] = temp # Overwrite.
self .sparse[temp] = self .sparse[x] # Overwrite.
# Since one element has been deleted, we
# decrement 'n' by 1.
self .n - = 1
# prints contents of set which are also content
# of dense[]
def print_set( self ):
for i in range ( self .n):
print ( self .dense[i], end = ' ' )
print ()
# A function to find intersection of two sets
def intersection( self , s):
# Capacity and max value of result set
iCap = min ( self .n, s.n)
iMaxVal = max (s.maxValue, self .maxValue)
# Create result set
result = SSet(iMaxVal, iCap)
# Find the smaller of two sets
# If this set is smaller
if self .n < s.n:
# Search every element of this set in 's'.
# If found, add it to result
for i in range ( self .n):
if s.search( self .dense[i]) ! = - 1 :
result.insert( self .dense[i])
else :
# Search every element of 's' in this set.
# If found, add it to result
for i in range (s.n):
if self .search(s.dense[i]) ! = - 1 :
result.insert(s.dense[i])
return result
# A function to find union of two sets
# Time Complexity-O(n1+n2)
def setUnion( self , set2):
uCap = self .n + set2.n
uMaxVal = max ( self .maxValue, set2.maxValue)
# Initialize an empty set to store the union
union_set = SSet(uMaxVal, uCap)
# Insert all elements from the first set into the union set
for i in range ( self .n):
union_set.insert( self .dense[i])
# Insert all elements from the second set into the union set
for i in range (set2.n):
union_set.insert(set2.dense[i])
return union_set
# Driver program if __name__ = = "__main__" :
# Create a set set1 with capacity 5 and max value 100
s1 = SSet( 100 , 5 )
# Insert elements into the set set1
s1.insert( 5 )
s1.insert( 3 )
s1.insert( 9 )
s1.insert( 10 )
# Printing the elements in the data structure
print ( "The elements in set1 are:" )
s1.print_set()
index = s1.search( 3 )
# 'index' variable stores the index of the number to be searched.
if index ! = - 1 : # 3 exists
print (f "\n3 is found at index {index} in set1" )
else : # 3 doesn't exist
print ( "\n3 doesn't exist in set1" )
# Delete 9 and print set1
s1.deletion( 9 )
s1.print_set()
# Create a set with capacity 6 and max value 1000
s2 = SSet( 1000 , 6 )
# Insert elements into the set
s2.insert( 4 )
s2.insert( 3 )
s2.insert( 7 )
s2.insert( 200 )
# Printing set 2
print ( "\nThe elements in set2 are:" )
s2.print_set()
# Printing the intersection of the two sets
intersect = s2.intersection(s1)
print ( "\nIntersection of set1 and set2" )
intersect.print_set()
# Printing the union of the two sets
unionset = s1.setUnion(s2)
print ( "\nUnion of set1 and set2" )
unionset.print_set()
# This code is contributed by amit_mangal_ |
using System;
public class SSet
{ private int [] sparse;
private int [] dense;
private int n;
private int capacity;
private int maxValue;
public SSet( int maxV, int cap)
{
sparse = new int [maxV + 1];
dense = new int [cap];
capacity = cap;
maxValue = maxV;
n = 0;
}
public int Search( int x)
{
if (x > maxValue)
return -1;
if (sparse[x] < n && dense[sparse[x]] == x)
return (sparse[x]);
return -1;
}
public void Insert( int x)
{
if (n == capacity)
return ;
if (x > maxValue)
return ;
int i = sparse[x];
if (i >= n || dense[i] != x)
{
dense[n] = x;
sparse[x] = n;
n++;
}
}
public void Deletion( int x)
{
if (x > maxValue)
return ;
int i = sparse[x];
if (i < n && dense[i] == x)
{
n--;
dense[i] = dense[n];
sparse[dense[n]] = i;
}
}
public void Print()
{
Console.Write( "Sparse Set: " );
for ( int i = 0; i < n; i++)
Console.Write(dense[i] + " " );
Console.WriteLine();
}
public void Clear()
{
n = 0;
}
public SSet Intersection(SSet s)
{
SSet result = new SSet(maxValue, capacity);
for ( int i = 0; i < n; i++)
{
if (s.Search(dense[i]) != -1)
result.Insert(dense[i]);
}
return result;
}
public SSet SetUnion(SSet s)
{
SSet result = new SSet(maxValue, capacity);
for ( int i = 0; i < n; i++)
result.Insert(dense[i]);
for ( int i = 0; i < s.n; i++)
result.Insert(s.dense[i]);
return result;
}
} public class Program
{ public static void Main()
{
// Create a set set1 with capacity 5 and max
// value 100
SSet s1 = new SSet(100, 5);
// Insert elements into the set set1
s1.Insert(5);
s1.Insert(3);
s1.Insert(9);
s1.Insert(10);
// Printing the elements in the data structure.
Console.WriteLine( "The elements in set1 are" );
s1.Print();
int index = s1.Search(3);
// 'index' variable stores the index of the number to
// be searched.
if (index != -1) // 3 exists
Console.WriteLine( "\n3 is found at index {0} in set1" , index);
else // 3 doesn't exist
Console.WriteLine( "\n3 doesn't exists in set1" );
// Delete 9 and print set1
s1.Deletion(9);
s1.Print();
// Create a set with capacity 6 and max value
// 1000
SSet s2 = new SSet(1000, 6);
// Insert elements into the set
s2.Insert(4);
s2.Insert(3);
s2.Insert(7);
s2.Insert(200);
// Printing set 2.
Console.WriteLine( "\nThe elements in set2 are" );
s2.Print();
// Printing the intersection of the two sets
SSet intersect = s2.Intersection(s1);
Console.WriteLine( "\nIntersection of set1 and set2" );
intersect.Print();
// Printing the union of the two sets
SSet unionset = s1.SetUnion(s2);
Console.WriteLine( "\nUnion of set1 and set2" );
unionset.Print();
}
} |
// A JavaScript program to implement Sparse Set and its operations // A structure to hold the three parameters required to // represent a sparse set. class SSet { // Constructor
constructor(maxV, cap) {
this .sparse = new Array(maxV+1).fill(0);
this .dense = new Array(cap).fill(0);
this .capacity = cap;
this .maxValue = maxV;
this .n = 0; // No elements initially
}
// If element is present, returns index of
// element in dense[]. Else returns -1.
search(x) {
// Searched element must be in range
if (x > this .maxValue)
return -1;
// The first condition verifies that 'x' is
// within 'n' in this set and the second
// condition tells us that it is present in
// the data structure.
if ( this .sparse[x] < this .n && this .dense[ this .sparse[x]] === x)
return this .sparse[x];
// Not found
return -1;
}
// Inserts a new element into set
insert(x) {
// Corner cases, x must not be out of
// range, dense[] should not be full and
// x should not already be present
if (x > this .maxValue)
return ;
if ( this .n >= this .capacity)
return ;
if ( this .search(x) !== -1)
return ;
// Inserting into array-dense[] at index 'n'.
this .dense[ this .n] = x;
// Mapping it to sparse[] array.
this .sparse[x] = this .n;
// Increment count of elements in set
this .n++;
}
// A function that deletes 'x' if present in this data
// structure, else it does nothing (just returns).
// By deleting 'x', we unset 'x' from this set.
deletion(x) {
const index = this .search(x);
// If x is not present
if (index === -1)
return ;
const temp = this .dense[ this .n-1]; // Take an element from end
this .dense[index] = temp; // Overwrite.
this .sparse[temp] = index; // Overwrite.
// Since one element has been deleted, we
// decrement 'n' by 1.
this .n--;
}
// prints contents of set which are also content
// of dense[]
print() {
console.log( this .dense.slice(0, this .n).join( ' ' ));
}
clear() { this .n = 0; }
// A function to find intersection of two sets
intersection(s) {
// Capacity and max value of result set
const iCap = Math.min( this .n, s.n);
const iMaxVal = Math.max(s.maxValue, this .maxValue);
// Create result set
const result = new SSet(iMaxVal, iCap);
// Find the smaller of two sets
// If this set is smaller
if ( this .n < s.n) {
// Search every element of this set in 's'.
// If found, add it to result
for (let i = 0; i < this .n; i++)
if (s.search( this .dense[i]) !== -1)
result.insert( this .dense[i]);
}
else {
// Search every element of 's' in this set.
// If found, add it to result
for (let i = 0; i < s.n; i++)
if ( this .search(s.dense[i]) !== -1)
result.insert(s.dense[i]);
}
return result;
}
// A function to find union of two sets
// Time Complexity-O(n1+n2)
setUnion(s) {
// Find capacity and maximum value for result
// set.
const uCap = s.n + this .n;
const uMaxVal = Math.max(s.maxValue, this .maxValue);
// Create result set
const result = new SSet(uMaxVal, uCap);
// Traverse the first set and insert all
// elements of it in result.
for (let i = 0; i < this .n; i++)
result.insert( this .dense[i]);
// Traverse the second set and insert all
// elements of it in result (Note that sparse
// set doesn't insert an entry if it is already
// present)
for (let i = 0; i < s.n; i++)
result.insert(s.dense[i]);
return result;
}
} // Driver program // Create a set set1 with capacity 5 and max // value 100 const s1 = new SSet(100, 5);
// Insert elements into the set set1 s1.insert(5); s1.insert(3); s1.insert(9); s1.insert(10); // Printing the elements in the data structure. console.log('The elements in set1 are ');
s1.print(); const index = s1.search(3); // ' index ' variable stores the index of the number to
// be searched. if (index !== -1) // 3 exists console.log(`\n3 is found at index ${index} in set1`);
else // 3 doesn' t exist
console.log( '\n3 doesn\'t exists in set1' );
// Delete 9 and print set1 s1.deletion(9); s1.print(); // Create a set with capacity 6 and max value // 1000 const s2 = new SSet(1000, 6);
// Insert elements into the set s2.insert(4); s2.insert(3); s2.insert(7); s2.insert(200); // Printing set 2. console.log( '\nThe elements in set2 are' );
s2.print(); // Printing the intersection of the two sets const intersect = s2.intersection(s1); console.log( '\nIntersection of set1 and set2' );
intersect.print(); // Printing the union of the two sets const unionset = s1.setUnion(s2); console.log( "\nUnion of set1 and set2" );
unionset.print() |
Output :
The elements in set1 are 5 3 9 10 3 is found at index 1 in set1 5 3 10 The elements in set2 are- 4 3 7 200 Intersection of set1 and set2 3 Union of set1 and set2 5 3 10 4 7 200
This is a C++ program that implements a sparse set and its operations. A sparse set is a data structure that is useful when the maximum value of the set is known, and the set contains only a small subset of the maximum value range. The sparse set uses two arrays, sparse and dense, to store the elements. The sparse array maps each element to its index in the dense array, and the dense array contains the actual elements of the set.
The program defines a class SSet to represent the sparse set. The constructor takes the maximum value of the set and the capacity of the dense array as input parameters. The search, insert, deletion, print, clear, intersection, and setUnion functions are defined for the class.
The search function takes an element as input and returns its index in the dense array if it is present, otherwise -1. The insert function takes an element as input and inserts it into the set if it is not already present and the set is not full. The deletion function takes an element as input and removes it from the set if it is present. The print function prints the elements of the set in the dense array. The clear function removes all elements from the set.
The intersection function takes another SSet object as input and returns a new SSet object that contains the intersection of the two sets. The setUnion function takes another SSet object as input and returns a new SSet object that contains the union of the two sets. The intersection and setUnion functions use the search and insert functions to perform the set operations.
Overall, this program provides a simple and efficient implementation of a sparse set and its operations.
Additional Operations:
The following operations are also efficiently implemented using sparse set. It outperforms all the solutions discussed here and bit vector based solution, under the assumptions that range and maximum number of elements are known.
union():
1) Create an empty sparse set, say result.
2) Traverse the first set and insert all elements of it in result.
3) Traverse the second set and insert all elements of it in result (Note that sparse set doesn’t insert an entry if it is already present)
4) Return result.
intersection():
1) Create an empty sparse set, say result.
2) Let the smaller of two given sets be first set and larger be second.
3) Consider the smaller set and search every element of it in second. If element is found, add it to result.
4) Return result.
A common use of this data structure is with register allocation algorithms in compilers, which have a fixed universe(the number of registers in the machine) and are updated and cleared frequently (just like- Q queries) during a single processing run.
References:
http://research.switch.com/sparse
http://codingplayground.blogspot.in/2009/03/sparse-sets-with-o1-insert-delete.html