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 = 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 

Approach: It is known that the N^th even number is given by 2 * K and the N^th odd number is given by 2 * K – 1. But since the even numbers are written after (N + 1) / 2 odd numbers here. Therefore, K^th 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 <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)

Auxiliary Space: O(1)