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: 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 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++

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// 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;
}

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Java

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// 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));
    }
}

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Python3

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# 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

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C#

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// 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

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PHP

Output:

8

Time Complexity: O(1)



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Improved By : shrikanth13, Code_Mech