Given two integers **N** and **K**, the task is to find the element at the **K ^{th}** 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 = 3Output:5

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

**Approach:** It is known that the **N ^{th}** even number is given by

**2 * K**and the

**N**odd number is given by

^{th}**2 * K – 1**. But since the even numbers are written after

**(N + 1) / 2**odd numbers here. Therfore,

**K**even number is given by

^{th}**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 <bits/stdc++.h>` `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

`<script>` `// Javascript implementation of the approach` `// Function to return the kth number` `// from the required sequence` `function` `kthNum(n, k)` `{` ` ` `// Count of odd integers` ` ` `// in the sequence` ` ` `var` `a = (n + 1) / 2;` ` ` `// kth number is even` ` ` `if` `(k > a)` ` ` `return` `(2 * (k - a));` ` ` `// It is odd` ` ` `return` `(2 * k - 1);` `}` `// Driver code` `var` `n = 7, k = 7;` `document.write(kthNum(n, k));` `</script>` |

**Output:**

6

**Time Complexity:** O(1)

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