# Queries to find kth smallest element and point update : Ordered Set in C++

• Difficulty Level : Hard
• Last Updated : 14 Apr, 2020

Given an array arr[] of size N and a set Q[][] containing M queries, the task is to execute the queries on the given array such that there can be two types of queries:

• Type 1: [i, x] – Update the element at ith index to x.
• Type 2: [k] – Find the kth smallest element in the array.

Examples:

Input: arr[] = {4, 3, 6, 2}, Q[][] = {{1, 2, 5}, {2, 3}, {1, 1, 7}, {2, 1}}
Output: 5 2
Explanation:
For the 1st query: arr[] = {4, 5, 6, 2}
For the 2nd query: 3rd smallest element would be 5.
For the 3rd query: arr[] = {7, 5, 6, 2}
For the 4th query: 1st smallest element would be 2.

Input: arr[] = {1, 0, 4, 2, 0}, Q[][] = {{1, 2, 1}, {2, 2}, {1, 4, 5}, {1, 3, 7}, {2, 1}, {2, 5}}
Output: 1 0 7

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: The naive approach for this problem is to update the ith element in an array in constant time and use sorting to find the Kth smallest element.

Time Complexity: O(M * (N * log(N))) where M is the number of queries and N is the size of the array.

Efficient Approach: The idea is to use a policy-based data structure similar to a set.

Here, a tree based container is used to store the array in the form of a sorted tree such that all the nodes to the left are smaller than the root and all the nodes to the right are greater than the root. The following are the properties of the data structure:

• It is indexed by maintaining node invariant where each node contains a count of nodes in its subtree.
• Every time we insert a new node or delete a node, we can maintain the invariant in O(logN) time by bubbling up to the root.
• So the count of the node in its left subtree gives the index of that node in sorted order because the value of every node of the left subtree is smaller than the parent node.

Therefore, the idea is to follow the following approach for each query:

1. Type 1: For this query, we update the ith element of the array. Therefore, we need to update the element both in the array and the data structure. In order to update the value in the tree container, the value arr[i] is found in the tree, deleted from the tree and the updated value is inserted back into the tree.
2. Type 2: In order to find the Kth smallest element, find_by_order(K – 1) is used on the tree as the data is a sorted data. This is similar to Binary Search operation on the sorted array.

Below is the implementation of the above approach:

 `// C++ implementation of the above approach`` ` `#include ``#include ``#include ``using` `namespace` `std;``using` `namespace` `__gnu_pbds;`` ` `// Defining the policy based Data Structure``typedef` `tree,``             ``null_type,``             ``less >,``             ``rb_tree_tag,``             ``tree_order_statistics_node_update>``    ``indexed_set;`` ` `// Elements in the array are not unique,``// so a pair is used to give uniqueness``// by incrementing cnt and assigning``// with array elements to insert in mySet``int` `cnt = 0;`` ` `// Variable to store the data in the``// policy based Data Structure``indexed_set mySet;`` ` `// Function to insert the elements``// of the array in mySet``void` `insert(``int` `n, ``int` `arr[])``{``    ``for` `(``int` `i = 0; i < n; i++) {``        ``mySet.insert({ arr[i], cnt });``        ``cnt++;``    ``}``}`` ` `// Function to update the value in``// the data structure``void` `update(``int` `x, ``int` `y)``{``    ``// Get the pointer of the element``    ``// in mySet which has to be updated``    ``auto` `it = mySet.lower_bound({ y, 0 });`` ` `    ``// Delete from mySet``    ``mySet.erase(it);`` ` `    ``// Insert the updated value in mySet``    ``mySet.insert({ x, cnt });``    ``cnt++;``}`` ` `// Function to find the K-th smallest``// element in the set``int` `get(``int` `k)``{``    ``// Find the pointer to the kth smallest element``    ``auto` `it = mySet.find_by_order(k - 1);``    ``return` `(it->first);``}`` ` `// Function to perform the queries on the set``void` `operations(``int` `arr[], ``int` `n,``                ``vector > query, ``int` `m)``{``    ``// To insert the element in mySet``    ``insert(n, arr);`` ` `    ``// Iterating through the queries``    ``for` `(``int` `i = 0; i < m; i++) {`` ` `        ``// Checking if the query is of type 1``        ``// or type 2``        ``if` `(query[i] == 1) {`` ` `            ``// The array is 0-indexed``            ``int` `j = query[i] - 1;``            ``int` `x = query[i];`` ` `            ``// Update the element in mySet``            ``update(x, arr[j]);`` ` `            ``// Update the element in the array``            ``arr[j] = x;``        ``}``        ``else` `{``            ``int` `K = query[i];`` ` `            ``// Print Kth smallest element``            ``cout << get(K) << endl;``        ``}``    ``}``}`` ` `// Driver code``int` `main()``{``    ``int` `n = 5, m = 6, arr[] = { 1, 0, 4, 2, 0 };`` ` `    ``vector > query = { { 1, 2, 1 },``                                   ``{ 2, 2 },``                                   ``{ 1, 4, 5 },``                                   ``{ 1, 3, 7 },``                                   ``{ 2, 1 },``                                   ``{ 2, 5 } };`` ` `    ``operations(arr, n, query, m);`` ` `    ``return` `0;``}`

Output:

```1
0
7
```

Time Complexity: Since every operation takes O(Log(N)) time and there are M queries, the overall time complexity is O(M * Log(N)).

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