Sparse Set

How to do the following operations efficiently if there are large number of queries for them.

  1. Insertion
  2. Deletion
  3. Searching
  4. 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

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 (positiion 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[].


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.

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 C++ implementation of above functions.

/* A C program to implement Sparse Set and its operations */
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[] */

    // 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
        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)
    if (n >= capacity)
    if (search(x) != -1)

    // Inserting into array-dense[] at index 'n'.
    dense[n] = x;

    // Mapping it to sparse[] array.
    sparse[x] = n;

    // Increment count of elements in set

// 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)

    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.

// prints contents of set which are also content
// of dense[]
void SSet::print()
    for (int i=0; i<n; i++)
        printf("%d ", dense[i]);

// 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 ([i]) != -1)
        // 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)

    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++)

    // 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++)

    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

    // Printing the elements in the data structure.
    printf("The elements in set1 are\n");

    int index =;

    //  '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

    // Create a set with capacity 6 and max value
    // 1000
    SSet s2(1000, 6);

    // Insert elements into the set

    // Printing set 2.
    printf("\nThe elements in set2 are\n");

    // Printing the intersection of the two sets
    SSet *intersect = s2.intersection(s1);
    printf("\nIntersection of set1 and set2\n");

    // Printing the union of the two sets
    SSet *unionset = s1.setUnion(s2);
    printf("\nUnion of set1 and set2\n");

    return 0;

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

Union of set1 and set2
5 3 10 4 7 200 

Additional Operations:
The following are 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.

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.

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.


This article is contributed by Rachit Belwariar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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