# Find Kth element in an array containing odd elements first and then even elements

• Difficulty Level : Basic
• Last Updated : 14 Jun, 2022

Given the length of an array of integers N and an integer K. The task is to modify the array in such a way that the array contains first all odd integers from 1 to N in ascending order, then all even integers from 1 to N in ascending order and then print the Kth element in the modified array.
Examples:

Input: N = 8, K = 5
Output:
The array will be {1, 3, 5, 7, 2, 4, 6, 8}
and the fifth element is 2.
Input: N = 7, K = 2
Output:

Naive approach: A simple approach is to store the odd numbers first, one by one till N, and then storing the even numbers one by one till N, and then printing the kth element.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the kth element``// in the modified array``int` `getNumber(``int` `n, ``int` `k)``{``    ``int` `arr[n];` `    ``int` `i = 0;` `    ``// First odd number``    ``int` `odd = 1;``    ``while` `(odd <= n) {` `        ``// Insert the odd number``        ``arr[i++] = odd;` `        ``// Next odd number``        ``odd += 2;``    ``}` `    ``// First even number``    ``int` `even = 2;``    ``while` `(even <= n) {` `        ``// Insert the even number``        ``arr[i++] = even;` `        ``// Next even number``        ``even += 2;``    ``}` `    ``// Return the kth element``    ``return` `arr[k - 1];``}` `// Driver code``int` `main()``{``    ``int` `n = 8, k = 5;``    ``cout << getNumber(n, k);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{` `// Function to return the kth element``// in the modified array``static` `int` `getNumber(``int` `n, ``int` `k)``{``    ``int` `[]arr = ``new` `int``[n];` `    ``int` `i = ``0``;` `    ``// First odd number``    ``int` `odd = ``1``;``    ``while` `(odd <= n)``    ``{` `        ``// Insert the odd number``        ``arr[i++] = odd;` `        ``// Next odd number``        ``odd += ``2``;``    ``}` `    ``// First even number``    ``int` `even = ``2``;``    ``while` `(even <= n)``    ``{` `        ``// Insert the even number``        ``arr[i++] = even;` `        ``// Next even number``        ``even += ``2``;``    ``}` `    ``// Return the kth element``    ``return` `arr[k - ``1``];``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``8``, k = ``5``;``    ``System.out.println(getNumber(n, k));``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of the approach` `# Function to return the kth element``# in the modified array``def` `getNumber(n, k):``    ``arr ``=` `[``0``] ``*` `n;` `    ``i ``=` `0``;` `    ``# First odd number``    ``odd ``=` `1``;``    ``while` `(odd <``=` `n):``        ` `        ``# Insert the odd number``        ``arr[i] ``=` `odd;``        ``i ``+``=` `1``;` `        ``# Next odd number``        ``odd ``+``=` `2``;` `    ``# First even number``    ``even ``=` `2``;``    ``while` `(even <``=` `n):``        ``# Insert the even number``        ``arr[i] ``=` `even;``        ``i ``+``=` `1``;` `        ``# Next even number``        ``even ``+``=` `2``;` `    ``# Return the kth element``    ``return` `arr[k ``-` `1``];` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``n ``=` `8``;``    ``k ``=` `5``;``    ``print``(getNumber(n, k));` `# This code is contributed by Rajput-Ji`

## C#

 `// C# implementation of the approach``using` `System;``    ` `class` `GFG``{` `// Function to return the kth element``// in the modified array``static` `int` `getNumber(``int` `n, ``int` `k)``{``    ``int` `[]arr = ``new` `int``[n];` `    ``int` `i = 0;` `    ``// First odd number``    ``int` `odd = 1;``    ``while` `(odd <= n)``    ``{` `        ``// Insert the odd number``        ``arr[i++] = odd;` `        ``// Next odd number``        ``odd += 2;``    ``}` `    ``// First even number``    ``int` `even = 2;``    ``while` `(even <= n)``    ``{` `        ``// Insert the even number``        ``arr[i++] = even;` `        ``// Next even number``        ``even += 2;``    ``}` `    ``// Return the kth element``    ``return` `arr[k - 1];``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `n = 8, k = 5;``    ``Console.WriteLine(getNumber(n, k));``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output:

`2`

Time Complexity: O(n)
Auxiliary Space: O(n), as extra space of size n is used

Efficient approach: Find the index where the first even element will be stored in the generated array. Now if the value of k is less than or equal to the index then the desired number will be k * 2 – 1 else the desired number will be (k – index) * 2
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the kth element``// in the modified array``int` `getNumber(``int` `n, ``int` `k)``{``    ``int` `pos;` `    ``// Finding the index from where the``    ``// even numbers will be stored``    ``if` `(n % 2 == 0) {``        ``pos = n / 2;``    ``}``    ``else` `{``        ``pos = (n / 2) + 1;``    ``}` `    ``// Return the kth element``    ``if` `(k <= pos) {``        ``return` `(k * 2 - 1);``    ``}``    ``else` `        ``return` `((k - pos) * 2);``}` `// Driver code``int` `main()``{``    ``int` `n = 8, k = 5;` `    ``cout << getNumber(n, k);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.io.*;` `class` `GFG``{``    ` `// Function to return the kth element``// in the modified array``static` `int` `getNumber(``int` `n, ``int` `k)``{``    ``int` `pos;` `    ``// Finding the index from where the``    ``// even numbers will be stored``    ``if` `((n % ``2``) == ``0``)``    ``{``        ``pos = n / ``2``;``    ``}``    ``else``    ``{``        ``pos = (n / ``2``) + ``1``;``    ``}` `    ``// Return the kth element``    ``if` `(k <= pos)``    ``{``        ``return` `(k * ``2` `- ``1``);``    ``}``    ``else``        ``return` `((k - pos) * ``2``);``}` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``int` `n = ``8``, k = ``5``;``    ``System.out.println (getNumber(n, k));``}``}` `// This code is contributed by @tushil.`

## Python3

 `# Python3 implementation of the approach` `# Function to return the kth element``# in the modified array``def` `getNumber(n, k) :` `    ``# Finding the index from where the``    ``# even numbers will be stored``    ``if` `(n ``%` `2` `=``=` `0``) :``        ``pos ``=` `n ``/``/` `2``;``    ` `    ``else` `:``        ``pos ``=` `(n ``/``/` `2``) ``+` `1``;` `    ``# Return the kth element``    ``if` `(k <``=` `pos) :``        ``return` `(k ``*` `2` `-` `1``);``        ` `    ``else` `:``        ``return` `((k ``-` `pos) ``*` `2``);` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:``    ``n ``=` `8``; k ``=` `5``;``    ` `    ``print``(getNumber(n, k));` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``        ` `// Function to return the kth element``// in the modified array``static` `int` `getNumber(``int` `n, ``int` `k)``{``    ``int` `pos;` `    ``// Finding the index from where the``    ``// even numbers will be stored``    ``if` `((n % 2) == 0)``    ``{``        ``pos = n / 2;``    ``}``    ``else``    ``{``        ``pos = (n / 2) + 1;``    ``}` `    ``// Return the kth element``    ``if` `(k <= pos)``    ``{``        ``return` `(k * 2 - 1);``    ``}``    ``else``        ``return` `((k - pos) * 2);``}` `// Driver code``static` `public` `void` `Main ()``{``    ``int` `n = 8, k = 5;``    ``Console.Write(getNumber(n, k));``}``}` `// This code is contributed by @ajit.`

## Javascript

 ``

Output:

`2`

Time Complexity: O(1)
Auxiliary Space: O(1)

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