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Find the kth element in the series generated by the given N ranges
  • Difficulty Level : Basic
  • Last Updated : 03 Jun, 2021

Given N non-overlapping ranges L[] and R[] where the every range starts after the previous range ends i.e. L[i] > R[i – 1] for all valid i. The task is to find the Kth element in the series which is formed after sorting all the elements in all the given ranges in ascending order.
Examples: 
 

Input: L[] = {1, 8, 21}, R[] = {4, 10, 23}, K = 6 
Output:
The generated series will be 1, 2, 3, 4, 8, 9, 10, 21, 22, 23 
And the 6th element is 9
Input: L[] = {2, 11, 31}, R[] = {7, 15, 43}, K = 13 
Output: 32 
 

 

Approach: The idea is to use binary search. An array total to store the number of integers that are present upto ith index, now with the help of this array find out the index in which the kth integer will lie. Suppose that index is j, now compute the position of the kth smallest integer in the interval L[j] to R[j] and find the kth smallest integer using binary search where low will be L[j] and high will be R[j].
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the kth element
// of the required series
int getKthElement(int n, int k, int L[], int R[])
{
    int l = 1;
    int h = n;
 
    // To store the number of integers that lie
    // upto the ith index
    int total[n + 1];
 
    total[0] = 0;
 
    // Compute the number of integers
    for (int i = 0; i < n; i++) {
        total[i + 1] = total[i] + (R[i] - L[i]) + 1;
    }
 
    // Stores the index, lying from 1
    // to n,
    int index = -1;
 
    // Using binary search, find the index
    // in which the kth element will lie
    while (l <= h) {
        int m = (l + h) / 2;
 
        if (total[m] > k) {
            index = m;
            h = m - 1;
        }
        else if (total[m] < k)
            l = m + 1;
        else {
            index = m;
            break;
        }
    }
 
    l = L[index - 1];
    h = R[index - 1];
 
    // Find the position of the kth element
    // in the interval in which it lies
    int x = k - total[index - 1];
 
    while (l <= h) {
        int m = (l + h) / 2;
 
        if ((m - L[index - 1]) + 1 == x) {
            return m;
        }
 
        else if ((m - L[index - 1]) + 1 > x)
            h = m - 1;
 
        else
            l = m + 1;
    }
}
 
// Driver code
int main()
{
    int L[] = { 1, 8, 21 };
    int R[] = { 4, 10, 23 };
    int n = sizeof(L) / sizeof(int);
 
    int k = 6;
 
    cout << getKthElement(n, k, L, R);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
     
// Function to return the kth element
// of the required series
static int getKthElement(int n, int k,
                         int L[], int R[])
{
    int l = 1;
    int h = n;
 
    // To store the number of integers that lie
    // upto the ith index
    int total[] = new int[n + 1];
 
    total[0] = 0;
 
    // Compute the number of integers
    for (int i = 0; i < n; i++)
    {
        total[i + 1] = total[i] +
                      (R[i] - L[i]) + 1;
    }
 
    // Stores the index, lying from 1
    // to n,
    int index = -1;
 
    // Using binary search, find the index
    // in which the kth element will lie
    while (l <= h)
    {
        int m = (l + h) / 2;
 
        if (total[m] > k)
        {
            index = m;
            h = m - 1;
        }
        else if (total[m] < k)
            l = m + 1;
        else
        {
            index = m;
            break;
        }
    }
 
    l = L[index - 1];
    h = R[index - 1];
 
    // Find the position of the kth element
    // in the interval in which it lies
    int x = k - total[index - 1];
 
    while (l <= h)
    {
        int m = (l + h) / 2;
 
        if ((m - L[index - 1]) + 1 == x)
        {
            return m;
        }
 
        else if ((m - L[index - 1]) + 1 > x)
            h = m - 1;
 
        else
            l = m + 1;
    }
    return k;
}
 
// Driver code
public static void main(String[] args)
{
    int L[] = { 1, 8, 21 };
    int R[] = { 4, 10, 23 };
    int n = L.length;
 
    int k = 6;
 
    System.out.println(getKthElement(n, k, L, R));
}
}
 
// This code is contributed by Code_Mech

Python3




# Python3 implementation of the approach
  
# Function to return the kth element
# of the required series
def getKthElement(n, k, L, R):
    l = 1
    h = n
  
    # To store the number of integers that lie
    # upto the ith index
    total=[0 for i in range(n + 1)]
  
    total[0] = 0
  
    # Compute the number of integers
    for i in range(n):
        total[i + 1] = total[i] + (R[i] - L[i]) + 1
  
    # Stores the index, lying from 1
    # to n,
    index = -1
  
    # Using binary search, find the index
    # in which the kth element will lie
    while (l <= h):
        m = (l + h) // 2
  
        if (total[m] > k):
            index = m
            h = m - 1
        elif (total[m] < k):
            l = m + 1
        else :
            index = m
            break
  
    l = L[index - 1]
    h = R[index - 1]
  
    # Find the position of the kth element
    # in the interval in which it lies
    x = k - total[index - 1]
  
    while (l <= h):
        m = (l + h) // 2
  
        if ((m - L[index - 1]) + 1 == x):
            return m
  
        elif ((m - L[index - 1]) + 1 > x):
            h = m - 1
  
        else:
            l = m + 1
 
# Driver code
 
L=[ 1, 8, 21]
R=[4, 10, 23]
n = len(L)
 
k = 6
 
print(getKthElement(n, k, L, R))
 
# This code is contributed by mohit kumar

C#




// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the kth element
// of the required series
static int getKthElement(int n, int k,
                        int[] L, int[] R)
{
    int l = 1;
    int h = n;
 
    // To store the number of integers that lie
    // upto the ith index
    int[] total = new int[n + 1];
 
    total[0] = 0;
 
    // Compute the number of integers
    for (int i = 0; i < n; i++)
    {
        total[i + 1] = total[i] +
                    (R[i] - L[i]) + 1;
    }
 
    // Stores the index, lying from 1
    // to n,
    int index = -1;
 
    // Using binary search, find the index
    // in which the kth element will lie
    while (l <= h)
    {
        int m = (l + h) / 2;
 
        if (total[m] > k)
        {
            index = m;
            h = m - 1;
        }
        else if (total[m] < k)
            l = m + 1;
        else
        {
            index = m;
            break;
        }
    }
 
    l = L[index - 1];
    h = R[index - 1];
 
    // Find the position of the kth element
    // in the interval in which it lies
    int x = k - total[index - 1];
 
    while (l <= h)
    {
        int m = (l + h) / 2;
 
        if ((m - L[index - 1]) + 1 == x)
        {
            return m;
        }
 
        else if ((m - L[index - 1]) + 1 > x)
            h = m - 1;
 
        else
            l = m + 1;
    }
    return k;
}
 
// Driver code
public static void Main()
{
    int[] L = { 1, 8, 21 };
    int[] R = { 4, 10, 23 };
    int n = L.Length;
 
    int k = 6;
 
    Console.WriteLine(getKthElement(n, k, L, R));
}
}
 
// This code is contributed by Code_Mech

PHP




<?php
// PHP implementation of the approach
 
// Function to return the kth element
// of the required series
function getKthElement($n, $k, $L, $R)
{
    $l = 1;
    $h = $n;
 
    // To store the number of integers that lie
    // upto the ith index
    $total = array();
 
    $total[0] = 0;
 
    // Compute the number of integers
    for ($i = 0; $i < $n; $i++)
    {
        $total[$i + 1] = $total[$i] +
                        ($R[$i] - $L[$i]) + 1;
    }
 
    // Stores the index, lying from 1
    // to n,
    $index = -1;
 
    // Using binary search, find the index
    // in which the kth element will lie
    while ($l <= $h)
    {
        $m = floor(($l + $h) / 2);
 
        if ($total[$m] > $k)
        {
            $index = $m;
            $h = $m - 1;
        }
        else if ($total[$m] < $k)
            $l = $m + 1;
        else
        {
            $index = $m;
            break;
        }
    }
 
    $l = $L[$index - 1];
    $h = $R[$index - 1];
 
    // Find the position of the kth element
    // in the interval in which it lies
    $x = $k - $total[$index - 1];
 
    while ($l <= $h)
    {
        $m = floor(($l + $h) / 2);
 
        if (($m - $L[$index - 1]) + 1 == $x)
        {
            return $m;
        }
 
        else if (($m - $L[$index - 1]) + 1 > $x)
            $h = $m - 1;
 
        else
            $l = $m + 1;
    }
}
 
// Driver code
$L = array( 1, 8, 21 );
$R = array( 4, 10, 23 );
$n = count($L);
 
$k = 6;
 
echo getKthElement($n, $k, $L, $R);
 
// This code is contributed by Ryuga
?>

Javascript




<script>
// Javascript implementation of the approach
 
// Function to return the kth element
// of the required series
function getKthElement(n,k,L,R)
{
    let l = 1;
    let h = n;
   
    // To store the number of integers that lie
    // upto the ith index
    let total = new Array(n + 1);
   
    total[0] = 0;
   
    // Compute the number of integers
    for (let i = 0; i < n; i++)
    {
        total[i + 1] = total[i] +
                      (R[i] - L[i]) + 1;
    }
   
    // Stores the index, lying from 1
    // to n,
    let index = -1;
   
    // Using binary search, find the index
    // in which the kth element will lie
    while (l <= h)
    {
        let m = Math.floor((l + h) / 2);
   
        if (total[m] > k)
        {
            index = m;
            h = m - 1;
        }
        else if (total[m] < k)
            l = m + 1;
        else
        {
            index = m;
            break;
        }
    }
   
    l = L[index - 1];
    h = R[index - 1];
   
    // Find the position of the kth element
    // in the interval in which it lies
    let x = k - total[index - 1];
   
    while (l <= h)
    {
        let m = Math.floor((l + h) / 2);
   
        if ((m - L[index - 1]) + 1 == x)
        {
            return m;
        }
   
        else if ((m - L[index - 1]) + 1 > x)
            h = m - 1;
   
        else
            l = m + 1;
    }
    return k;
}
 
// Driver code
let L = [1, 8, 21 ];
let R = [ 4, 10, 23 ];
let n = L.length;
let k = 6;
 
document.write(getKthElement(n, k, L, R));
 
// This code is contributed by patel2127
</script>
Output: 
9

 

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