Order statistic tree using fenwick tree (BIT)

Given an array of integers with limited range (0 to 1000000). We need to implement an Order statistic tree using fenwick tree.
It should support four operations: Insert, Delete, Select and Rank. Here n denotes the size of Fenwick tree and q denotes number of queries.
Each query should be one of the following 4 operations.

  • insertElement(x) – Insert element x into Fenwick tree, with O(log n) worst case time complexity
  • deleteElement(x) – Delete element x from fenwick tree, with O(log n) worse case time complexity
  • findKthSmallest(k) – Find the k-th smallest element stored in the tree, with O(log n * log n) worst case time complexity
  • findRank(x) – Find the rank of element x in the tree, i.e. its index in the sorted list of elements of the tree, with O(log n) time complexity

Prerequisite : Binary Indexed Tree or Fenwick Tree

The idea is to create a BIT of size with maximum limit. We insert an element in BIT using it as an index. When we insert an element x, we increment values of all ancestors of x by 1. To delete an element, we decrement values of ancestors by 1. We basically call standard function update() of BIT for both insert and delete. To find rank, we simply call standard function sum() of BIT. To find k-th smallest element, we do binary search in BIT.

[sourcecode language=”CPP”]
// C++ program to find rank of an element
// and k-th smallest element.
#include <bits/stdc++.h>
using namespace std;

const int MAX_VAL = 1000001;

/* Updates element at index ‘i’ of BIT. */
void update(int i, int add, vector<int>& BIT)
{
while (i > 0 && i < BIT.size())
{
BIT[i] += add;
i = i + (i & (-i));
}
}

/* Returns cumulative sum of all elements of
fenwick tree/BIT from start upto and
including element at index ‘i’. */
int sum(int i, vector<int>& BIT)
{
int ans = 0;
while (i > 0)
{
ans += BIT[i];
i = i – (i & (-i));
}

return ans;
}

// Returns lower bound for k in BIT.
int findKthSmallest(int k, vector<int> &BIT)
{
// Do binary search in BIT[] for given
// value k.
int l = 0;
int h = BIT.size();
while (l < h)
{
int mid = (l + h) / 2;
if (k <= sum(mid, BIT))
h = mid;
else
l = mid+1;
}

return l;
}

// Insert x into BIT. We masically increment
// rank of all elements greater than x.
void insertElement(int x, vector<int> &BIT)
{
update(x, 1, BIT);
}

// Delete x from BIT. We masically decreases
// rank of all elements greater than x.
void deleteElement(int x, vector<int> &BIT)
{
update(x, -1, BIT);
}

// Returns rank of element. We basically
// return sum of elements from start to
// index x.
int findRank(int x, vector<int> &BIT)
{
return sum(x, BIT);
}

// Driver code
int main()
{
vector<int> BIT(MAX_VAL);
insertElement(20, BIT);
insertElement(50, BIT);
insertElement(30, BIT);
insertElement(40, BIT);

cout << "2nd Smallest element is "
<< findKthSmallest(2, BIT) << endl;

cout << "Rank of 40 is "
<< findRank(40, BIT) << endl;

deleteElement(40, BIT);

cout << "Rank of 50 is "
<< findRank(50, BIT) << endl;

return 0;
}
[/sourcecode]
Output:

2nd Smallest element is 30
Rank of 40 is 3
Rank of 50 is 3

This article is contributed by Kapil Lamba. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.



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Improved By : kapillamba4



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