Given an empty set initially and a number of queries on it, each possibly of the following types:

**Insert**– Insert a new element ‘x’.**Delete**– Delete an existing element ‘x’.**Median**– Print the median element of the numbers currently in the set

**Example:**

Input : Insert 1 Insert 4 Insert 7 Median Output : The first three queries should insert 1, 4 and 7 into an empty set. The fourth query should return 4 (median of 1, 4, 7).

For expository purpose, we assume the following, but these assumptions are not the limitations of the method discussed here:

1. At any instance, all elements are distinct, that is, none of them occurs more than once.

2. The ‘Median’ query is made only when there are odd number of elements in the set.( We will need to make two queries on our segment tree, in case of even numbers )

3. The elements in the set range from 1 to +10^6.

**Method 1 (Naive)**

In naive implementation, we can do first two queries in O(1), but the last query in O(max_elem), where max_elem is the maximum element of all times (including deleted elements).

Let us assume an array **count[]** (of size 10^6 + 1) to maintain the count of each element in the subset. Following are simple and self explanatory algorithms for the 3 queries:

**Insert x query:**

count[x]++; if (x > max_elem) max_elem = x; n++;

**Delete x query:**

if (count[x] > 0) count[x]--; n--;

**Median query:**

sum = 0; i = 0; while( sum <= n / 2 ) { i++; sum += count[i]; } median = i; return median;

Illustration of array count[], representing the set {1, 4, 7, 8, 9}, the median element is ‘7’:

0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 |

The ‘Median’ query intends to find the (n + 1)/2 th ‘1’ in the array, in this case, 3rd ‘1’; now we do the same using a segment tree.

**Method 2(Using Segment Tree)**

We make a segment tree to store sum of intervals, where an interval [a, b] represents the number of elements present in the set, currently, in the range [a, b]. For example, if we consider the above example, query( 3, 7) returns 2, query(4, 4) returns 1, query(5, 5) returns 0.

Insert and delete queries are simple and both can be implemented using function update(int x, int diff) (adds ‘diff’ at index ‘x’)

**Algorithm**

// adds ‘diff’ at index ‘x’update(node, a, b, x, diff)// If leaf node If a == b and a == x segmentTree[node] += diff // If non-leaf node and x lies in its range If x is in [a, b] // Update children recursively update(2*node, a, (a + b)/2, x, diff) update(2*node + 1, (a + b)/2 + 1, b, x, diff) // Update node segmentTree[node] = segmentTree[2 * node] + segmentTree[2 * node + 1]

The above recursive function runs in **O( log( max_elem ) )** ( in this case max_elem is 10^6) and used for both insertion and deletion with the following calls:

- Insert ‘x’ is done using update(1, 0, 10^6, x, 1). Note that root of tree is passed, start index is passed as 0 and end index as 10^6 so that all ranges that have x are updated.
- Delete ‘x’ is done using update(1, 0, 10^6, x, -1). Note that root of tree is passed, start index is passed as 0 and end index as 10^6 so that all ranges that have x are updated.

Now, the function to find the index with kth ‘1’, where ‘k’ in this case will always be (n + 1) / 2, this is going to work a lot like binary search, you can think of it as a recursive binary search function on a segment tree.

Let’s take an example to understand, our set currently has elements { 1, 4, 7, 8, 9 }, and hence is represented by the following segment tree.

The image is taken from http://visualgo.net/segmenttree

If we are at a non-leaf node, we are sure that it has both children, we see if the left child has more or equal number of one’s as ‘k’, if yes, we are sure our index lies in the left subtree, otherwise, if left subtree has less number of 1’s than k, then we are sure that our index lies in the right subtree. We do this recursively to reach our index and from there, we return it.

**Algorithm**

1.findKth(node, a, b, k) 2. If a != b 3. If segmentTree[ 2 * node ] >= k 4. return findKth(2*node, a, (a + b)/2, k) 5. else 6. return findKth(2*node + 1, (a + b)/2 + 1, b, k - segmentTree[ 2 * node ]) 7. else 8. return a

The above recursive function runs in **O( log(max_elem) )**.

// A C++ program to implement insert, delete and // median queries using segment tree #include<bits/stdc++.h> #define maxn 3000005 #define max_elem 1000000 using namespace std; // A global array to store segment tree. // Note: Since it is global, all elements are 0. int segmentTree[maxn]; // Update 'node' and its children in segment tree. // Here 'node' is index in segmentTree[], 'a' and // 'b' are starting and ending indexes of range stored // in current node. // 'diff' is the value to be added to value 'x'. void update(int node, int a, int b, int x, int diff) { // If current node is a leaf node if (a == b && a == x ) { // add 'diff' and return segmentTree[node] += diff; return ; } // If current node is non-leaf and 'x' is in its // range if (x >= a && x <= b) { // update both sub-trees, left and right update(node*2, a, (a + b)/2, x, diff); update(node*2 + 1, (a + b)/2 + 1, b, x, diff); // Finally update current node segmentTree[node] = segmentTree[node*2] + segmentTree[node*2 + 1]; } } // Returns k'th node in segment tree int findKth(int node, int a, int b, int k) { // non-leaf node, will definitely have both // children; left and right if (a != b) { // If kth element lies in the left subtree if (segmentTree[node*2] >= k) return findKth(node*2, a, (a + b)/2, k); // If kth one lies in the right subtree return findKth(node*2 + 1, (a + b)/2 + 1, b, k - segmentTree[node*2]); } // if at a leaf node, return the index it stores // information about return (segmentTree[node])? a : -1; } // insert x in the set void insert(int x) { update(1, 0, max_elem, x, 1); } // delete x from the set void delet(int x) { update(1, 0, max_elem, x, -1); } // returns median element of the set with odd // cardinality only int median() { int k = (segmentTree[1] + 1) / 2; return findKth(1, 0, max_elem, k); } // Driver code int main() { insert(1); insert(4); insert(7); cout << "Median for the set {1,4,7} = " << median() << endl; insert(8); insert(9); cout << "Median for the set {1,4,7,8,9} = " << median() << endl; delet(1); delet(8); cout << "Median for the set {4,7,9} = " << median() << endl; return 0; }

Output:

Median for the set {1,4,7} = 4 Median for the set {1,4,7,8,9} = 7 Median for the set {4,7,9} = 7

**Conclusion:**

All three queries run in **O( log(max_elem) )**, in this case max_elem = 10^6, so log(max_elem) is approximately equal to 20.

The segment tree uses **O( max_elem )** space.

If the delete query wasn’t there, the problem could have also been done with a famous algorithm here.

This article is contributed by **Saumye Malhotra ** .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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