# Find the Kth position element of the given sequence

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: 5
The required sequence is 1, 3, 5, 7, 9, 2, 4, 6, 8 and 10.

Input: N = 7, K = 7
Output: 6

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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. Therfore, 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 `

Output:

```6
```

Time Complexity: O(1)

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