Find k-th smallest element in given n ranges

Difficulty Level : Medium
Last Updated : 02 Mar, 2021

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++

// 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;
}

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<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

Python3

# Python3 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

C#

// C# implementation to solve k queries
// for given n ranges
using System;
using System.Collections;
using System.Collections.Generic;
  
class GFG{
  
// Structure to store the
// start and end point
public class Interval
{
    public int s;
    public int e;
     
    public Interval(int a, int b)
    {
        s = a;
        e = b;
    }
};
 
class sortHelper : IComparer
{
    int IComparer.Compare(object a, object b)
    {
        Interval first = (Interval)a;
        Interval second = (Interval)b;
        return first.s - second.s;
    }
}
 
// Function to find Kth smallest number in
// a Vector of merged intervals
static int kthSmallestNum(ArrayList merged, int k)
{
    int n = merged.Count;
     
    // Traverse merged.get( )o find
    // Kth smallest element using Linear search.
    for(int j = 0; j < n; j++)
    {
        if (k <= Math.Abs(((Interval)merged[j]).e -
                          ((Interval)merged[j]).s + 1))
            return (((Interval)merged[j]).s + k - 1);
  
        k = k - Math.Abs(((Interval)merged[j]).e -
                         ((Interval)merged[j]).s + 1);
    }
  
    if (k != 0)
        return -1;
         
    return 0;
}
  
// To combined both type of ranges,
// overlapping as well as non-overlapping.
static ArrayList mergeIntervals(ArrayList merged,
                                Interval []arr, int n)
{
     
    // Sorting intervals according to start
    // time
    Array.Sort(arr, new sortHelper());
     
    // Merging all intervals into merged
    merged.Add((Interval)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 = (Interval)merged[merged.Count - 1];
        Interval curr = arr[i];
         
        if ((curr.s >= prev.s && curr.s <= prev.e) &&
            (curr.e > prev.e))
        {
            ((Interval)merged[merged.Count - 1]).e = ((Interval)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())
    ArrayList merged = new ArrayList();
     
    merged = mergeIntervals(merged, arr, n);
     
    // Processing all queries on merged
    // intervals
    for(int i = 0; i < q; i++)
        Console.WriteLine(kthSmallestNum(
            merged, query[i]));
}
}
 
// This code is contributed by pratham76

Output:

 6
 -1

Time Complexity : O(nlog(n))
Auxiliary Space: O(n)

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