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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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