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# 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 kth smallest element from the range [1, n] after deleting all the odd numbers from the range.
Examples:

Input: n = 8, k = 3
Output:
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 3rd 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 kth smallest element will always be 2 * k.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``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

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## Javascript

 ``

Output:

`8`

Time Complexity: O(1)

Auxiliary Space: O(1)