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# Find the Kth position element of the given sequence

• Last Updated : 07 Mar, 2022

Given two integers N and K, the task is to find the element at the Kth position if all odd numbers from 1 to N are written down in increasing order followed by all the even numbers from 1 to N in increasing order.
Examples:

Input: N = 10, K = 3
Output:
The required sequence is 1, 3, 5, 7, 9, 2, 4, 6, 8 and 10.
Input: N = 7, K = 7
Output:

Approach: It is known that the Nth even number is given by 2 * K and the Nth odd number is given by 2 * K – 1. But since the even numbers are written after (N + 1) / 2 odd numbers here. Therefore, Kth even number is given by 2 * (K – (N + 1) / 2) and the odd numbers will remain the same as 2 * K – 1
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the kth number``// from the required sequence``int` `kthNum(``int` `n, ``int` `k)``{` `    ``// Count of odd integers``    ``// in the sequence``    ``int` `a = (n + 1) / 2;` `    ``// kth number is even``    ``if` `(k > a)``        ``return` `(2 * (k - a));` `    ``// It is odd``    ``return` `(2 * k - 1);``}` `// Driver code``int` `main()``{``    ``int` `n = 7, k = 7;` `    ``cout << kthNum(n, k);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{` `// Function to return the kth number``// from the required sequence``static` `int` `kthNum(``int` `n, ``int` `k)``{` `    ``// Count of odd integers``    ``// in the sequence``    ``int` `a = (n + ``1``) / ``2``;` `    ``// kth number is even``    ``if` `(k > a)``        ``return` `(``2` `* (k - a));` `    ``// It is odd``    ``return` `(``2` `* k - ``1``);``}` `// Driver code``public` `static` `void` `main(String []args)``{``    ``int` `n = ``7``, k = ``7``;` `    ``System.out.println(kthNum(n, k));``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation of the approach` `# Function to return the kth number``# from the required sequence``def` `kthNum(n, k) :` `    ``# Count of odd integers``    ``# in the sequence``    ``a ``=` `(n ``+` `1``) ``/``/` `2``;` `    ``# kth number is even``    ``if` `(k > a) :``        ``return` `(``2` `*` `(k ``-` `a));` `    ``# It is odd``    ``return` `(``2` `*` `k ``-` `1``);` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``n ``=` `7``; k ``=` `7``;` `    ``print``(kthNum(n, k));` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;``    ` `class` `GFG``{` `// Function to return the kth number``// from the required sequence``static` `int` `kthNum(``int` `n, ``int` `k)``{` `    ``// Count of odd integers``    ``// in the sequence``    ``int` `a = (n + 1) / 2;` `    ``// kth number is even``    ``if` `(k > a)``        ``return` `(2 * (k - a));` `    ``// It is odd``    ``return` `(2 * k - 1);``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``int` `n = 7, k = 7;` `    ``Console.WriteLine(kthNum(n, k));``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output:

`6`

Time Complexity: O(1)

Auxiliary Space: O(1)

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