# Find the Kth position element of the given sequence

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the

DSA Self Paced Courseat a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please referComplete Interview Preparation Course.In case you wish to attend

live classeswith experts, please referDSA Live Classes for Working ProfessionalsandCompetitive Programming Live for Students.

**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. Therefore,

**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)