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 = 3Output: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 = 4Output:

**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++

`// 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

`// 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

`# 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#

`// 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` `// 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` `?>` |

## Javascript

`<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>` |

**Output:**

8

**Time Complexity:** O(1)

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