Introduction
The problem statement is: Given an input array ‘input[]’ of size ‘n’ and ‘q’ queries. Every query has three arguments, ‘l’, ‘r’ and ‘k’. We need to print the Kth smallest element in the range from l to r. l and r mark the start and end of the subarray, respectively.
Example
Input
INPUT[] = {3, 4, 0, 6, 8, 1}
Q = [4, 5, 2]
Output
8
Explanation
The 2nd smallest element in the subarray {8, 1} is 8.
Input
INPUT[] = {1, 5, 9, 0, 9, 1, 8, 3, 3, 2}
Q = [0, 5, 3]
Output
1
Explanation
Clearly, the 3rd smallest element in the subarray {1, 5, 9, 0, 9, 1} is 1.
Also see, Euclid GCD Algorithm
Solution Approach
Naive Approach
For any query(l, r, k), we first create a temporary array ‘temp’. We copy the elements of the array from i = l to r into temp. Then we sort temp. Finally, we print temp[k  1] as the answer. This approach is quite easy to implement.
The time complexity of this approach is O(NlogN) as we are sorting. The space complexity is O(N) as we are using temp as an auxiliary array.
Efficient Approach
Idea
The idea is to build a merge sort tree(segment tree) in which every node contains an array of sorted elements of the subarray that is represented by that node. For more clarity, refer to the image attached below. For querying, the same traditional methodology is followed. However, everything is discussed in the next section.
Algorithm
 Declare a global 2D array ‘seg’, that refers to the merge sort tree. As already mentioned, every node contains a sorted subarray, ranging from ‘low’ to ‘high’.

Define a function buildST(idx, low, high, input), that builds the merge sort tree recursively:
 Understanding the parameters: A node is defined on the range ‘low’ to ‘high’. The node's data is stored in seg[idx] or, essentially, the node = seg[idx]. ‘input’ is the input array.
 Base case: when low = high, this means, that the range consists of only one element. Hence, simply set seg[i] = input[low].
 Else, calculate mid = (low + high) / 2.
 Recursively, call ‘buildST(2 * idx + 1, low, mid, input)’ and buildST(2 * idx + 2, mid + 1, high, input) to build the left and right subtrees. If the current node is seg[idx], the left node is seg[2 * idx + 1] and the right node is seg[2 * idx + 2]. This is an important property of merge sort trees.
 After recursively building the left and right subtrees, the values of left and right nodes are used to build the current node. FYI, the left and right node store sorted subarrays in a range [low, mid] and [mid + 1, high], which are merged to construct a new bigger subarray that is stored in seg[idx], i.e. the current node.

Define a wrapper function ‘query(l, r, k, n)’ as:
 ‘l’ is the start, and ‘r’ is the end of the query range. ‘k’ denotes that the Kth smallest number in the given range has to be returned. ‘n’ is the total number of elements in the ‘input’ array.
 This function calls ‘queryHelper(idx, l, r, low, high)’, which does all the heavy lifting and returns a sorted subarray in the range [l, r]. ‘idx’ points to the node corresponding to the range [low, high], a query range is bounded by [l, r].

This function is defined as:
 When the node lies completely inside the range [l, r] return seg[idx].
 When the node lies completely outside the range [l, r], return an empty array.
 Else, recursively call for the left child i.e. queryHelper(2 * idx + 1, l, r, low, mid) and queryHelper(2 * idx + 2, l, r, mid + 1, high), store the answers in ‘left’ and ‘right’ arrays, respectively, merge them and return the merger array.
 The wrapper function makes a call to queryHelper(0, l, r, 0, n  1), stores the answer of the call into an array and returns the element at the (k  1)th index.
C++ Code implementation
#include <bits/stdc++.h>
using namespace std;
// Globally declared
vector<vector<int>> seg;
// This function mergers two sorted arrays
vector<int> merge(vector<int> &a1, vector<int> &a2)
{
int n = a1.size(), m = a2.size();
int i = 0, j = 0;
vector<int> res;
while (i < n and j < m)
{
if (a1[i] < a2[j])
{
res.push_back(a1[i]);
i += 1;
}
else
{
res.push_back(a2[j]);
j += 1;
}
}
while (i < n)
{
res.push_back(a1[i++]);
}
while (j < m)
{
res.push_back(a2[j++]);
}
return res;
}
// This function builds the segment tree
void buildST(int idx, int low, int high, vector<int> &input)
{
// Base case: when there is a single elment in the range [LOW, HIGH]
if (low == high)
{
seg[idx].push_back({input[low]});
return;
}
int mid = (low + high) / 2;
// Recursive calls to build the left and right subtrees
buildST(2 * idx + 1, low, mid, input);
buildST(2 * idx + 2, mid + 1, high, input);
/*
The current node's value is calculated by merging
the values of the left and the right nodes.
*/
seg[idx] = merge(seg[2 * idx + 1], seg[2 * idx + 2]);
}
vector<int> queryHelper(int idx, int l, int r, int low, int high)
{
// When a node is completely inside
if (low >= l and high <= r)
{
return seg[idx];
}
// When a node is completely outside
if (r < low or l > high)
{
return vector<int>();
}
int mid = (low + high) / 2;
/*
Fetch sorted subarray from the left and the right subtrees
and merge them to get a sorted array in the range [l, r]
*/
vector<int> left = queryHelper(2 * idx + 1, l, r, low, mid);
vector<int> right = queryHelper(2 * idx + 2, l, r, mid + 1, high);
return merge(left, right);
}
/*
Wrapper function which triggers queryHelper,
returns the Kth smallest element in the given range [l, r]
*/
int query(int l, int r, int k, int n)
{
vector<int> res = queryHelper(0, l, r, 0, n  1);
return res[k  1];
}
int main()
{
vector<int> arr{3, 4, 0, 6, 8, 1};
int n = arr.size();
// Setting up globals
seg.clear();
seg.resize(4 * n);
// Building the segment tree
buildST(0, 0, n  1, arr);
// Querying and returning the answer
cout << query(4, 5, 2, n) << endl;
}
Output
8
Complexity Analysis
Time complexity: O(Q * NlogN)
For a single query, the most that we can go down in the merge sort tree is the height. The height of the tree ~logN, and at every call, we perform a merge operation that takes O(N) time in the worst case. If there are Q queries to answer, the overall time complexity becomes O(Q * NlogN).
Space Complexity: O(N ^ 2)
We need some auxiliary space to store the algorithm's merge sort tree that we build. The size of the merge sort tree is 4 * N, and every element in the tree is an array itself. The size of this array at max can be N because the nodes in the merge sort tree store a subarray of the given input array. Therefore the overall space complexity is O(N ^ 2).
Read More  Time Complexity of Sorting Algorithms