Given an array of n numbers, the task is to answer the following queries:

kthSmallest(start, end, k) : Find the K^{th}smallest number in the range from array index 'start' to 'end'.

Examples:

Input : arr[] = {3, 2, 5, 1, 8, 9| Query 1: start = 2, end = 5, k = 2 Query 2: start = 1, end = 6, k = 4 Output : 2 5Explanation:[2, 5, 1, 8] represents the range from 2 to 5 and 2 is the 2^{nd}smallest number in the range[3, 2, 5, 1, 8, 9] represents the range from 1 to 6 and 5 is the 4^{th}smallest number in the range

The key idea is to build a Segment Tree with a vector at every node and the vector contains all the elements of the sub-range in a sorted order. And if we observe this segment tree structure this is somewhat similar to the tree formed during the merge sort algorithm(that is why it is called merge sort tree)

We use same implementation as discussed in Merge Sort Tree (Smaller or equal elements in given row range)

Firstly, we maintain a vector of pairs where each pair {value, index} is such that first element of pair represents the element of the input array and the second element of the pair represents the index at which it occurs.

Now we sort this vector of pairs on the basis of the first element of each pair.

After this we build a Merge Sort Tree where each node has a vector of indices in the sorted range.

When we have to answer a query we find if the K^{th} smallest number lies in the left sub-tree or in the right sub-tree. The idea is to use two binary searches and find the number of elements in the left sub-tree such that that the indices lie within the given query range.

Let the number of such indices be M.

If M>=K, it means we will be able to find the K^{th} smallest Number in the left sub-tree thus we call on the left sub-tree.

Else the K^{th} smallest number lies in the right sub-tree but this time we don’t have to look for the K ^{th} smallest number as we already have first M smallest numbers of the range in the left sub-tree thus we should look for the remaining part ie the (K-M)^{th} number in the right sub-tree.

This is the Index of K^{th} smallest number the value at this index is the required number.

`// CPP program to implement k-th order statistics` `#include <bits/stdc++.h>` `using` `namespace` `std;` ` ` `const` `int` `MAX = 1000;` ` ` `// Constructs a segment tree and stores tree[]` `void` `buildTree(` `int` `treeIndex, ` `int` `l, ` `int` `r, ` ` ` `vector<pair<` `int` `, ` `int` `> > &a, vector<` `int` `> tree[])` `{` ` ` ` ` `/* l => start of range,` ` ` `r => ending of a range` ` ` `treeIndex => index in the Segment Tree/Merge ` ` ` `Sort Tree */` ` ` ` ` `/* leaf node */` ` ` `if` `(l == r) {` ` ` `tree[treeIndex].push_back(a[l].second);` ` ` `return` `;` ` ` `}` ` ` ` ` `int` `mid = (l + r) / 2;` ` ` ` ` `/* building left subtree */` ` ` `buildTree(2 * treeIndex, l, mid, a, tree);` ` ` ` ` `/* building left subtree */` ` ` `buildTree(2 * treeIndex + 1, mid + 1, r, a, tree);` ` ` ` ` `/* merging left and right child in sorted order */` ` ` `merge(tree[2 * treeIndex].begin(), ` ` ` `tree[2 * treeIndex].end(),` ` ` `tree[2 * treeIndex + 1].begin(), ` ` ` `tree[2 * treeIndex + 1].end(),` ` ` `back_inserter(tree[treeIndex]));` `}` ` ` `// Returns the Kth smallest number in query range` `int` `queryRec(` `int` `segmentStart, ` `int` `segmentEnd, ` ` ` `int` `queryStart, ` `int` `queryEnd, ` `int` `treeIndex,` ` ` `int` `K, vector<` `int` `> tree[])` `{` ` ` `/*` ` ` `segmentStart => start of a Segment,` ` ` `segmentEnd => ending of a Segment,` ` ` `queryStart => start of a query range,` ` ` `queryEnd => ending of a query range,` ` ` `treeIndex => index in the Segment ` ` ` `Tree/Merge Sort Tree,` ` ` `K => kth smallest number to find */` ` ` ` ` `if` `(segmentStart == segmentEnd) ` ` ` `return` `tree[treeIndex][0];` ` ` ` ` `int` `mid = (segmentStart + segmentEnd) / 2;` ` ` ` ` `// finds the last index in the segment ` ` ` `// which is <= queryEnd` ` ` `int` `last_in_query_range = ` ` ` `(upper_bound(tree[2 * treeIndex].begin(),` ` ` `tree[2 * treeIndex].end(),` ` ` `queryEnd)` ` ` `- tree[2 * treeIndex].begin());` ` ` ` ` `// finds the first index in the segment` ` ` `// which is >= queryStart` ` ` `int` `first_in_query_range = ` ` ` `(lower_bound(tree[2 * treeIndex].begin(),` ` ` `tree[2 * treeIndex].end(),` ` ` `queryStart)` ` ` `- tree[2 * treeIndex].begin());` ` ` ` ` `int` `M = last_in_query_range - first_in_query_range;` ` ` ` ` `if` `(M >= K) {` ` ` ` ` `// Kth smallest is in left subtree,` ` ` `// so recursively call left subtree for Kth ` ` ` `// smallest number` ` ` `return` `queryRec(segmentStart, mid, queryStart, ` ` ` `queryEnd, 2 * treeIndex, K, tree);` ` ` `}` ` ` ` ` `else` `{` ` ` ` ` `// Kth smallest is in right subtree,` ` ` `// so recursively call right subtree for the ` ` ` `// (K-M)th smallest number` ` ` `return` `queryRec(mid + 1, segmentEnd, queryStart,` ` ` `queryEnd, 2 * treeIndex + 1, K - M, tree);` ` ` `}` `}` ` ` `// A wrapper over query()` `int` `query(` `int` `queryStart, ` `int` `queryEnd, ` `int` `K, ` `int` `n,` ` ` `vector<pair<` `int` `, ` `int` `> > &a, vector<` `int` `> tree[])` `{` ` ` ` ` `return` `queryRec(0, n - 1, queryStart - 1, queryEnd - 1, ` ` ` `1, K, tree);` `}` ` ` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = { 3, 2, 5, 1, 8, 9 };` ` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]);` ` ` ` ` `// vector of pairs of form {element, index}` ` ` `vector<pair<` `int` `, ` `int` `> > v;` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `v.push_back(make_pair(arr[i], i));` ` ` `}` ` ` ` ` `// sort the vector` ` ` `sort(v.begin(), v.end());` ` ` ` ` `// Construct segment tree in tree[]` ` ` `vector<` `int` `> tree[MAX];` ` ` `buildTree(1, 0, n - 1, v, tree);` ` ` ` ` `// Answer queries` ` ` `// kSmallestIndex hold the index of the kth smallest number` ` ` `int` `kSmallestIndex = query(2, 5, 2, n, v, tree);` ` ` `cout << arr[kSmallestIndex] << endl;` ` ` ` ` `kSmallestIndex = query(1, 6, 4, n, v, tree);` ` ` `cout << arr[kSmallestIndex] << endl;` ` ` ` ` `return` `0;` `}` |

**Output:**

2 5

Thus, we can get the K^{th} smallest number query in range L to R, in O(n(logn)^{2}) by building the merge sort tree on indices.

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