# Find k-th smallest element in given n ranges

Given n and q, i.e, the number of ranges and number of queries, find the kth smallest element for each query (assume k>1).Print the value of kth smallest element if it exists, else print -1.

Examples :

```Input : arr[] = {{1, 4}, {6, 8}}
queries[] = {2, 6, 10};
Output : 2
7
-1
After combining the given ranges, the numbers
become 1 2 3 4 6 7 8. As here 2nd element is 2,
so we print 2. As 6th element is 7, so we print
7 and as 10th element doesn't exist, so we
print -1.

Input : arr[] = {{2, 6}, {5, 7}}
queries[] = {5, 8};
Output : 6
-1
After combining the given ranges, the numbers
become 2 3 4 5 6 7. As here 5th element is 6,
so we print 6 and as 8th element doesn't exist,
so we print -1.
```

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

The idea is to first Prerequisite : Merge Overlapping Intervals and keep all intervals sorted in ascending order of start time. After merging in an array merged[], we use linear search to find kth smallest element. Below is the implementation of the above approach :

## C++

 `// C++ implementation to solve k queries ` `// for given n ranges ` `#include ` `using` `namespace` `std; ` ` `  `// Structure to store the ` `// start and end point ` `struct` `Interval ` `{ ` `    ``int` `s; ` `    ``int` `e; ` `}; ` ` `  `// Comparison function for sorting ` `bool` `comp(Interval a, Interval b) ` `{ ` `    ``return` `a.s < b.s; ` `} ` ` `  `// Function to find Kth smallest number in a vector ` `// of merged intervals ` `int` `kthSmallestNum(vector merged, ``int` `k) ` `{ ` `    ``int` `n = merged.size(); ` ` `  `    ``// Traverse merged[] to find ` `    ``// Kth smallest element using Linear search. ` `    ``for` `(``int` `j = 0; j < n; j++) ` `    ``{ ` `        ``if` `(k <= ``abs``(merged[j].e - ` `                     ``merged[j].s + 1)) ` `            ``return` `(merged[j].s + k - 1); ` ` `  `        ``k = k - ``abs``(merged[j].e - ` `                     ``merged[j].s + 1); ` `    ``} ` ` `  `    ``if` `(k) ` `        ``return` `-1; ` `} ` ` `  `// To combined both type of ranges, ` `// overlapping as well as non-overlapping. ` `void` `mergeIntervals(vector &merged, ` `                 ``Interval arr[], ``int` `n) ` `{ ` `    ``// Sorting intervals according to start ` `    ``// time ` `    ``sort(arr, arr + n, comp); ` ` `  `    ``// Merging all intervals into merged ` `    ``merged.push_back(arr); ` `    ``for` `(``int` `i = 1; i < n; i++) ` `    ``{ ` `        ``// To check if starting point of next ` `        ``// range is lying between the previous ` `        ``// range and ending point of next range ` `        ``// is greater than the Ending point ` `        ``// of previous range then update ending ` `        ``// point of previous range by ending ` `        ``// point of next range. ` `        ``Interval prev = merged.back(); ` `        ``Interval curr = arr[i]; ` `        ``if` `((curr.s >= prev.s && ` `             ``curr.s <= prev.e) && ` `            ``(curr.e > prev.e)) ` ` `  `            ``merged.back().e = curr.e; ` ` `  `        ``else` `        ``{ ` `            ``// If starting point of next range ` `            ``// is greater than the ending point ` `            ``// of previous range then store next range ` `            ``// in merged[]. ` `            ``if` `(curr.s > prev.e) ` `                ``merged.push_back(curr); ` `        ``} ` `    ``} ` `} ` ` `  `// Driver\'s Function ` `int` `main() ` `{ ` `    ``Interval arr[] = {{2, 6}, {4, 7}}; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr); ` `    ``int` `query[] = {5, 8}; ` `    ``int` `q = ``sizeof``(query)/``sizeof``(query); ` ` `  `    ``// Merge all intervals into merged[] ` `    ``vectormerged; ` `    ``mergeIntervals(merged, arr, n); ` ` `  `    ``// Processing all queries on merged ` `    ``// intervals ` `    ``for` `(``int` `i = 0; i < q; i++) ` `        ``cout << kthSmallestNum(merged, query[i]) ` `             ``<< endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation to solve k queries ` `// for given n ranges ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Structure to store the ` `// start and end point ` `static` `class` `Interval ` `{ ` `    ``int` `s; ` `    ``int` `e; ` `    ``Interval(``int` `a,``int` `b) ` `    ``{ ` `        ``s = a; ` `        ``e = b; ` `    ``} ` `}; ` `static` `class` `Sortby ``implements` `Comparator  ` `{  ` `    ``// Comparison function for sorting ` `    ``public` `int` `compare(Interval a, Interval b) ` `    ``{ ` `        ``return` `a.s - b.s; ` `    ``} ` `}  ` ` `  `// Function to find Kth smallest number in a Vector ` `// of merged intervals ` `static` `int` `kthSmallestNum(Vector merged, ``int` `k) ` `{ ` `    ``int` `n = merged.size(); ` ` `  `    ``// Traverse merged.get( )o find ` `    ``// Kth smallest element using Linear search. ` `    ``for` `(``int` `j = ``0``; j < n; j++) ` `    ``{ ` `        ``if` `(k <= Math.abs(merged.get(j).e - ` `                    ``merged.get(j).s + ``1``)) ` `            ``return` `(merged.get(j).s + k - ``1``); ` ` `  `        ``k = k - Math.abs(merged.get(j).e - ` `                    ``merged.get(j).s + ``1``); ` `    ``} ` ` `  `    ``if` `(k != ``0``) ` `        ``return` `-``1``; ` `    ``return` `0``; ` `} ` ` `  `// To combined both type of ranges, ` `// overlapping as well as non-overlapping. ` `static` `Vector mergeIntervals(Vector merged, ` `                ``Interval arr[], ``int` `n) ` `{ ` `    ``// Sorting intervals according to start ` `    ``// time ` `    ``Arrays.sort(arr, ``new` `Sortby()); ` ` `  `    ``// Merging all intervals into merged ` `    ``merged.add(arr[``0``]); ` `    ``for` `(``int` `i = ``1``; i < n; i++) ` `    ``{ ` `        ``// To check if starting point of next ` `        ``// range is lying between the previous ` `        ``// range and ending point of next range ` `        ``// is greater than the Ending point ` `        ``// of previous range then update ending ` `        ``// point of previous range by ending ` `        ``// point of next range. ` `        ``Interval prev = merged.get(merged.size() - ``1``); ` `        ``Interval curr = arr[i]; ` `        ``if` `((curr.s >= prev.s && ` `            ``curr.s <= prev.e) && ` `            ``(curr.e > prev.e)) ` ` `  `            ``merged.get(merged.size()-``1``).e = curr.e; ` ` `  `        ``else` `        ``{ ` `            ``// If starting point of next range ` `            ``// is greater than the ending point ` `            ``// of previous range then store next range ` `            ``// in merged.get(.)         if (curr.s > prev.e) ` `                ``merged.add(curr); ` `        ``} ` `    ``} ` `    ``return` `merged; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``Interval arr[] = {``new` `Interval(``2``, ``6``), ``new` `Interval(``4``, ``7``)}; ` `    ``int` `n = arr.length; ` `    ``int` `query[] = {``5``, ``8``}; ` `    ``int` `q = query.length; ` ` `  `    ``// Merge all intervals into merged.get())  ` `    ``Vector merged = ``new` `Vector(); ` `    ``merged=mergeIntervals(merged, arr, n); ` ` `  `    ``// Processing all queries on merged ` `    ``// intervals ` `    ``for` `(``int` `i = ``0``; i < q; i++) ` `        ``System.out.println( kthSmallestNum(merged, query[i])); ` `} ` `} ` ` `  `// This code is contributed by Arnab Kundu `

## Python3

 `# Python implementation to solve k queries ` `# for given n ranges ` ` `  `# Structure to store the ` `# start and end point ` `class` `Interval: ` `    ``def` `__init__(``self``, s, e): ` `        ``self``.s ``=` `s ` `        ``self``.e ``=` `e ` ` `  `# Function to find Kth smallest number in a vector ` `# of merged intervals ` `def` `kthSmallestNum(merged: ``list``, k: ``int``) ``-``> ``int``: ` `    ``n ``=` `len``(merged) ` ` `  `    ``# Traverse merged[] to find ` `    ``# Kth smallest element using Linear search. ` `    ``for` `j ``in` `range``(n): ` `        ``if` `k <``=` `abs``(merged[j].e ``-` `merged[j].s ``+` `1``): ` `            ``return` `merged[j].s ``+` `k ``-` `1` ` `  `        ``k ``=` `k ``-` `abs``(merged[j].e ``-` `merged[j].s ``+` `1``) ` ` `  `    ``if` `k: ` `        ``return` `-``1` ` `  `# To combined both type of ranges, ` `# overlapping as well as non-overlapping. ` `def` `mergeIntervals(merged: ``list``, arr: ``list``, n: ``int``): ` ` `  `    ``# Sorting intervals according to start ` `    ``# time ` `    ``arr.sort(key ``=` `lambda` `a: a.s) ` ` `  `    ``# Merging all intervals into merged ` `    ``merged.append(arr[``0``]) ` `    ``for` `i ``in` `range``(``1``, n): ` ` `  `        ``# To check if starting point of next ` `        ``# range is lying between the previous ` `        ``# range and ending point of next range ` `        ``# is greater than the Ending point ` `        ``# of previous range then update ending ` `        ``# point of previous range by ending ` `        ``# point of next range. ` `        ``prev ``=` `merged[``-``1``] ` `        ``curr ``=` `arr[i] ` `        ``if` `curr.s >``=` `prev.s ``and` `curr.s <``=` `prev.e ``and``\ ` `          ``curr.e > prev.e: ` `            ``merged[``-``1``].e ``=` `curr.e ` `        ``else``: ` ` `  `            ``# If starting point of next range ` `            ``# is greater than the ending point ` `            ``# of previous range then store next range ` `            ``# in merged[]. ` `            ``if` `curr.s > prev.e: ` `                ``merged.append(curr) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``arr ``=` `[Interval(``2``, ``6``), Interval(``4``, ``7``)] ` `    ``n ``=` `len``(arr) ` `    ``query ``=` `[``5``, ``8``] ` `    ``q ``=` `len``(query) ` ` `  `    ``# Merge all intervals into merged[] ` `    ``merged ``=` `[] ` `    ``mergeIntervals(merged, arr, n) ` ` `  `    ``# Processing all queries on merged ` `    ``# intervals ` `    ``for` `i ``in` `range``(q): ` `        ``print``(kthSmallestNum(merged, query[i])) ` ` `  `# This code is contributed by ` `# sanjeev2552 `

Output:

``` 6
-1
```

Time Complexity : O(nlog(n))

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : sanjeev2552, andrew1234

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.