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.



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++

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// C++ implementation to solve k queries
// for given n ranges
#include <bits/stdc++.h>
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<Interval> 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<Interval> &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[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.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[0]);
    int query[] = {5, 8};
    int q = sizeof(query)/sizeof(query[0]);
  
    // Merge all intervals into merged[]
    vector<Interval>merged;
    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;
}

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Java














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// 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<Interval>  


{  


    // 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<Interval> 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<Interval> mergeIntervals(Vector<Interval> 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<Interval> merged = new Vector<Interval>(); 


    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 



















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Python3














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# 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 



















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Output:


 6
 -1



Time Complexity : O(nlog(n)) 


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