Find k-th smallest element in given n ranges
Last Updated :
12 Apr, 2023
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++
#include <bits/stdc++.h>
using namespace std;
struct Interval
{
int s;
int e;
};
bool comp(Interval a, Interval b)
{
return a.s < b.s;
}
int kthSmallestNum(vector<Interval> merged, int k)
{
int n = merged.size();
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;
}
void mergeIntervals(vector<Interval> &merged,
Interval arr[], int n)
{
sort(arr, arr + n, comp);
merged.push_back(arr[0]);
for ( int i = 1; i < n; i++)
{
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 (curr.s > prev.e)
merged.push_back(curr);
}
}
}
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]);
vector<Interval>merged;
mergeIntervals(merged, arr, n);
for ( int i = 0; i < q; i++)
cout << kthSmallestNum(merged, query[i])
<< endl;
return 0;
}
|
Java
import java.util.*;
class GFG
{
static class Interval
{
int s;
int e;
Interval( int a, int b)
{
s = a;
e = b;
}
};
static class Sortby implements Comparator<Interval>
{
public int compare(Interval a, Interval b)
{
return a.s - b.s;
}
}
static int kthSmallestNum(Vector<Interval> merged, int k)
{
int n = merged.size();
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 ;
}
static Vector<Interval> mergeIntervals(Vector<Interval> merged,
Interval arr[], int n)
{
Arrays.sort(arr, new Sortby());
merged.add(arr[ 0 ]);
for ( int i = 1 ; i < n; i++)
{
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
{
merged.add(curr);
}
}
return merged;
}
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;
Vector<Interval> merged = new Vector<Interval>();
merged=mergeIntervals(merged, arr, n);
for ( int i = 0 ; i < q; i++)
System.out.println( kthSmallestNum(merged, query[i]));
}
}
|
Python3
class Interval:
def __init__( self , s, e):
self .s = s
self .e = e
def kthSmallestNum(merged: list , k: int ) - > int :
n = len (merged)
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
def mergeIntervals(merged: list , arr: list , n: int ):
arr.sort(key = lambda a: a.s)
merged.append(arr[ 0 ])
for i in range ( 1 , n):
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 curr.s > prev.e:
merged.append(curr)
if __name__ = = "__main__" :
arr = [Interval( 2 , 6 ), Interval( 4 , 7 )]
n = len (arr)
query = [ 5 , 8 ]
q = len (query)
merged = []
mergeIntervals(merged, arr, n)
for i in range (q):
print (kthSmallestNum(merged, query[i]))
|
C#
using System;
using System.Collections;
using System.Collections.Generic;
class GFG{
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;
}
}
static int kthSmallestNum(ArrayList merged, int k)
{
int n = merged.Count;
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;
}
static ArrayList mergeIntervals(ArrayList merged,
Interval []arr, int n)
{
Array.Sort(arr, new sortHelper());
merged.Add((Interval)arr[0]);
for ( int i = 1; i < n; i++)
{
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
{
merged.Add(curr);
}
}
return merged;
}
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;
ArrayList merged = new ArrayList();
merged = mergeIntervals(merged, arr, n);
for ( int i = 0; i < q; i++)
Console.WriteLine(kthSmallestNum(
merged, query[i]));
}
}
|
Javascript
<script>
class Interval
{
constructor(a,b)
{
this .s=a;
this .e=b;
}
}
function kthSmallestNum(merged,k)
{
let n = merged.length;
for (let j = 0; j < n; j++)
{
if (k <= Math.abs(merged[j].e -
merged[j].s + 1))
return (merged[j].s + k - 1);
k = k - Math.abs(merged[j].e -
merged[j].s + 1);
}
if (k != 0)
return -1;
return 0;
}
function mergeIntervals(merged,arr,n)
{
arr.sort( function (a,b){ return a.s-b.s;});
merged.push(arr[0]);
for (let i = 1; i < n; i++)
{
let prev = merged[merged.length - 1];
let curr = arr[i];
if ((curr.s >= prev.s &&
curr.s <= prev.e) &&
(curr.e > prev.e))
merged[merged.length-1].e = curr.e;
else
{
merged.push(curr);
}
}
return merged;
}
let arr=[ new Interval(2, 6), new Interval(4, 7)];
let n = arr.length;
let query = [5, 8];
let q = query.length;
let merged = [];
merged=mergeIntervals(merged, arr, n);
for (let i = 0; i < q; i++)
document.write( kthSmallestNum(merged, query[i])+ "<br>" );
</script>
|
Time Complexity : O(nlog(n) + qn)
Auxiliary Space: O(n)
Efficient Approach using binary search:
The first step is same as the above approach. Sort the range intervals and then merge the overlapping intervals.
Now instead of searching linearly in merged ranges we can search using binary search if we store the prefix sum of number of elements in each range in a set.
For example: For the range { {1 4}, {6 8}, {9 10} } the number of elements in each range are : 4,3,2 respectively. Hence, prefix sum set would store {4 7 9}.
We can use the fact that the numbers are sorted to our advantage and use binary search to look for kth smallest element.
For kth smallest element we just have to find the lower_bound of k in set. the corresponding index will give us the index of the merged interval in which the required element is stored.
This works much faster for large number of queries and large N. Since the search time for each query is reduce to O(logn) from O(n).
Below is the implementation of the above approach :
C++
#include <bits/stdc++.h>
using namespace std;
vector<vector< int >> mergeIntervals(vector<vector< int >>&range){
int n=range.size();
vector<vector< int >>fin;
for ( int i=0;i<n-1;i++){
if (range[i][1] >= range[i+1][0]){
range[i+1][0]=min(range[i][0],range[i+1][0]);
range[i+1][1]=max(range[i][1],range[i+1][1]);
}
else {
fin.push_back(range[i]);
}
}
fin.push_back(range[n-1]);
return fin;
}
vector< int >kthSmallestNum(vector<vector< int >>&range, vector< int >queries){
sort(range.begin(),range.end());
vector<vector< int >>merged=mergeIntervals(range);
set< int >s;
int cumsum=0;
for ( auto cur_range:merged){
int num_ele=cur_range[1]-cur_range[0]+1;
cumsum+=num_ele;
s.insert(cumsum);
}
vector< int >fin;
for ( auto q:queries){
auto it=s.lower_bound(q);
if (it==s.end())fin.push_back(-1);
else if (it==s.begin()){
fin.push_back(merged[0][0]+q-1);
}
else {
int prevele=*prev(it);
int kth=q-prevele;
int idx=distance(s.begin(),it);
fin.push_back(merged[idx][0]+kth-1);
}
}
return fin;
}
int main()
{
vector<vector< int >>range = {{1, 4}, {6, 8}};
int n =range.size();
vector< int >queries = {2, 6, 10};
int q = queries.size();
vector< int >ans=kthSmallestNum(range, queries);
for ( auto it:ans)
cout<<it<< " " ;
return 0;
}
|
Java
import java.util.*;
public class Main {
public static List<Integer> kthSmallestNum(List<List<Integer>> range, List<Integer> queries) {
Collections.sort(range, Comparator.comparingInt(a -> a.get( 0 )));
List<List<Integer>> merged = mergeIntervals(range);
NavigableSet<Integer> s = new TreeSet<>();
int cumsum = 0 ;
for (List<Integer> cur_range : merged) {
int num_ele = cur_range.get( 1 ) - cur_range.get( 0 ) + 1 ;
cumsum += num_ele;
s.add(cumsum);
}
List<Integer> fin = new ArrayList<>();
for ( int q : queries) {
Integer tail = s.ceiling(q);
if (tail == null ) {
fin.add(- 1 );
} else {
int idx = s.headSet(tail, true ).size() - 1 ;
int prevele = idx == 0 ? 0 : s.headSet(tail, false ).last();
int kth = q - prevele;
fin.add(merged.get(idx).get( 0 ) + kth - 1 );
}
}
return fin;
}
private static List<List<Integer>> mergeIntervals(List<List<Integer>> range) {
int n = range.size();
List<List<Integer>> fin = new ArrayList<>();
for ( int i = 0 ; i < n - 1 ; i++) {
if (range.get(i).get( 1 ) >= range.get(i + 1 ).get( 0 )) {
range.get(i + 1 ).set( 0 , Math.min(range.get(i).get( 0 ), range.get(i + 1 ).get( 0 )));
range.get(i + 1 ).set( 1 , Math.max(range.get(i).get( 1 ), range.get(i + 1 ).get( 1 )));
} else {
fin.add(range.get(i));
}
}
fin.add(range.get(n - 1 ));
return fin;
}
public static void main(String[] args) {
List<List<Integer>> range = Arrays.asList(Arrays.asList( 1 , 4 ), Arrays.asList( 6 , 8 ));
List<Integer> queries = Arrays.asList( 2 , 6 , 10 );
List<Integer> ans = kthSmallestNum(range, queries);
for ( int it : ans) {
System.out.print(it + " " );
}
}
}
|
Python3
from bisect import bisect_left
from typing import List
def merge_intervals(intervals: List [ List [ int ]]) - > List [ List [ int ]]:
n = len (intervals)
fin = []
for i in range (n - 1 ):
if intervals[i][ 1 ] > = intervals[i + 1 ][ 0 ]:
intervals[i + 1 ][ 0 ] = min (intervals[i][ 0 ], intervals[i + 1 ][ 0 ])
intervals[i + 1 ][ 1 ] = max (intervals[i][ 1 ], intervals[i + 1 ][ 1 ])
else :
fin.append(intervals[i])
fin.append(intervals[n - 1 ])
return fin
def kth_smallest_num(intervals: List [ List [ int ]], queries: List [ int ]) - > List [ int ]:
intervals.sort()
merged = merge_intervals(intervals)
s = set ()
cumsum = 0
for cur_range in merged:
num_ele = cur_range[ 1 ] - cur_range[ 0 ] + 1
cumsum + = num_ele
s.add(cumsum)
ans = []
for q in queries:
it = bisect_left( sorted (s), q)
if it = = len (s):
ans.append( - 1 )
elif it = = 0 :
ans.append(merged[ 0 ][ 0 ] + q - 1 )
else :
prevele = sorted (s)[it - 1 ]
kth = q - prevele
idx = sorted (s).index(prevele)
ans.append(merged[idx + 1 ][ 0 ] + kth - 1 )
return ans
if __name__ = = '__main__' :
intervals = [[ 1 , 4 ], [ 6 , 8 ]]
queries = [ 2 , 6 , 10 ]
ans = kth_smallest_num(intervals, queries)
print (ans)
|
Javascript
function mergeIntervals(range) {
const n = range.length;
const fin = [];
for (let i = 0; i < n - 1; i++) {
if (range[i][1] >= range[i + 1][0]) {
range[i + 1][0] = Math.min(range[i][0], range[i + 1][0]);
range[i + 1][1] = Math.max(range[i][1], range[i + 1][1]);
} else {
fin.push(range[i]);
}
}
fin.push(range[n - 1]);
return fin;
}
function kthSmallestNum(range, queries) {
range.sort((a, b) => a[0] - b[0]);
const merged = mergeIntervals(range);
const s = new Set();
let cumsum = 0;
for (const cur_range of merged) {
const num_ele = cur_range[1] - cur_range[0] + 1;
cumsum += num_ele;
s.add(cumsum);
}
const fin = [];
for (const q of queries) {
const it = s.values();
let tail = it.next().value;
while (tail !== undefined && tail < q) {
tail = it.next().value;
}
if (tail === undefined) {
fin.push(-1);
} else {
const idx = Array.from(s.values()).indexOf(tail) - 1;
const prevele = idx === -1 ? 0 : Array.from(s.values())[idx];
const kth = q - prevele;
fin.push(merged[idx + 1][0] + kth - 1);
}
}
return fin;
}
const range = [[1, 4], [6, 8]];
const queries = [2, 6, 10];
const ans = kthSmallestNum(range, queries);
console.log(ans.join( " " ));
|
C#
using System;
using System.Collections.Generic;
class Program
{
static List<List< int >> MergeIntervals(List<List< int >> intervals)
{
int n = intervals.Count;
List<List< int >> fin = new List<List< int >>();
for ( int i = 0; i < n - 1; i++)
{
if (intervals[i][1] >= intervals[i + 1][0])
{
intervals[i + 1][0] = Math.Min(intervals[i][0], intervals[i + 1][0]);
intervals[i + 1][1] = Math.Max(intervals[i][1], intervals[i + 1][1]);
}
else
{
fin.Add(intervals[i]);
}
}
fin.Add(intervals[n - 1]);
return fin;
}
static List< int > KthSmallestNum(List<List< int >> intervals, List< int > queries)
{
intervals.Sort((x, y) => x[0].CompareTo(y[0]));
List<List< int >> merged = MergeIntervals(intervals);
SortedSet< int > s = new SortedSet< int >();
int cumsum = 0;
foreach (List< int > cur_range in merged)
{
int num_ele = cur_range[1] - cur_range[0] + 1;
cumsum += num_ele;
s.Add(cumsum);
}
List< int > ans = new List< int >();
foreach ( int q in queries)
{
int [] sorted_s = new List< int >(s).ToArray();
int it = Array.BinarySearch(sorted_s, q);
if (it < 0)
{
it = ~it;
}
if (it == s.Count)
{
ans.Add(-1);
}
else if (it == 0)
{
ans.Add(merged[0][0] + q - 1);
}
else
{
int prevele = sorted_s[it - 1];
int kth = q - prevele;
int idx = new List< int >(s).IndexOf(prevele);
ans.Add(merged[idx + 1][0] + kth - 1);
}
}
return ans;
}
static void Main()
{
List<List< int >> intervals = new List<List< int >> { new List< int > { 1, 4 }, new List< int > { 6, 8 } };
List< int > queries = new List< int > { 2, 6, 10 };
List< int > ans = KthSmallestNum(intervals, queries);
foreach ( int num in ans)
{
Console.Write(num + " " );
}
}
}
|
Time Complexity : O(nlog(n) + qlog(n))
Auxiliary Space: O(n)
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