Find kth smallest number in range [1, n] when all the odd numbers are deleted

Given two integers n and k, the task is to find the k^th smallest element from the range [1, n] after deleting all the odd numbers from the range.
Examples: 
 

Input: n = 8, k = 3 
Output: 6 
After deleting all the odd numbers from the range [1, 8] 
2, 4, 6 and 8 are the only numbers left and 6 is the 3^rd smallest.
Input: n = 8, k = 4 
Output: 
 

Approach: Since all odd numbers are removed so now only even numbers are left i.e. 2, 4, 6, 8, ..... 
Now, the k^th smallest element will always be 2 * k.
Below is the implementation of the above approach:
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the kth smallest
// element from the range [1, n] after
// removing all the odd elements
int kthSmallest(int n, int k)
{
    return (2 * k);
}

// Driver code
int main()
{
    int n = 8, k = 4;
    cout << kthSmallest(n, k);

    return 0;
}

// Java implementation of the approach
class GFG {

    // Function to return the kth smallest
    // element from the range [1, n] after
    // removing all the odd elements
    static int kthSmallest(int n, int k)
    {
        return (2 * k);
    }

    // Driver code
    public static void main(String args[])
    {
        int n = 8, k = 4;
        System.out.print(kthSmallest(n, k));
    }
}

# Python3 implementation of the approach 

# Function to return the kth smallest 
# element from the range [1, n] after 
# removing all the odd elements 
def kthSmallest(n, k):
    return 2 * k

# Driver code
n = 8; k = 4
print(kthSmallest(n, k))

# This code is contributed 
# by Shrikant13

// C# implementation of the approach
using System;
class GFG {

    // Function to return the kth smallest
    // element from the range [1, n] after
    // removing all the odd elements
    static int kthSmallest(int n, int k)
    {
        return (2 * k);
    }

    // Driver code
    public static void Main()
    {
        int n = 8, k = 4;
        Console.WriteLine(kthSmallest(n, k));
    }
}

// This code is contributed by Code_Mech

<?php
// PHP implementation of the approach

// Function to return the kth smallest
// element from the range [1, n] after
// removing all the odd elements
function kthSmallest($n, $k)
{
    return (2 * $k);
}

// Driver code
$n = 8; $k = 4;
echo(kthSmallest($n, $k));

// This code is contributed
// by Code_Mech
?>

<script>
// Javascript implementation of the approach

// Function to return the kth smallest
// element from the range [1, n] after
// removing all the odd elements
function kthSmallest(n, k)
{
    return (2 * k);
}

// Driver code
var n = 8, k = 4;
document.write(kthSmallest(n, k));

// This code is contributed by noob2000.
</script>

8

Time Complexity: O(1)

Auxiliary Space: O(1)