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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 ``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;` `    ``// 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;` `    ``// 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

 ` ``\$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

 ``
Output:
`9`

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