Kth element in permutation of first N natural numbers having all even numbers placed before odd numbers in increasing order

Given two integers N and K, the task is to find the K^th element in the permutation of first N natural numbers arranged such that all the even numbers appear before the odd numbers in increasing order.

Examples :

Input: N = 10, K = 3  
Output: 6
Explanation:
The required permutation is {2, 4, 6, 8, 10, 1, 3, 5, 7, 9}.
The 3^rd number in the permutation is 6.

Input: N = 5, K = 4
Output: 3
Explanation:
The required permutation is {2, 4, 1, 3, 5}.
The 4^th number in the permutation is 3.

Naive Approach: The simplest approach to solve the problem is to generate the required permutation of first N natural numbers and then traverse the permutation to find the K^th element present in it.
Follow the steps below to solve the problem:

• Initialize an array, say V[] of size N., to store the required sequence.
• Insert all even numbers less than or equal to N into V[].
• Then, insert all odd numbers less than or equal to N into V[].
• After forming the array, print the value of V[K – 1] as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the K-th element
// in the required permutation
void findKthElement(int N, int K)
{
    // Stores the required permutation
    vector<int> v;

    // Insert all the even numbers
    // less than or equal to N
    for (int i = 1; i <= N; i++) {
        if (i % 2 == 0) {
            v.push_back(i);
        }
    }

    // Now, insert all odd numbers
    // less than or equal to N
    for (int i = 1; i <= N; i++) {
        if (i % 2 != 0) {
            v.push_back(i);
        }
    }

    // Print the Kth element
    cout << v[K - 1];
}

// Driver Code
int main()
{
    int N = 10, K = 3;
    findKthElement(N, K);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG
{

  // Function to find the K-th element
  // in the required permutation
  static void findKthElement(int N, int K)
  {

    // Stores the required permutation
    ArrayList<Integer> v = new ArrayList<>();

    // Insert all the even numbers
    // less than or equal to N
    for (int i = 1; i <= N; i++) {
      if (i % 2 == 0) {
        v.add(i);
      }
    }

    // Now, insert all odd numbers
    // less than or equal to N
    for (int i = 1; i <= N; i++) {
      if (i % 2 != 0) {
        v.add(i);
      }
    }

    // Print the Kth element
    System.out.println(v.get(K - 1));
  }

  // Driver code
  public static void main(String[] args)
  {

    int N = 10, K = 3;

    // functions call
    findKthElement(N, K);
  }
}

// This code is contributed by Kingash.

Python3

# python 3 program for the above approach

# Function to find the K-th element
# in the required permutation
def findKthElement(N, K):

    # Stores the required permutation
    v = []

    # Insert all the even numbers
    # less than or equal to N
    for i in range(1, N + 1):
        if (i % 2 == 0):
            v.append(i)

    # Now, insert all odd numbers
    # less than or equal to N
    for i in range(1, N + 1):
        if (i % 2 != 0):
            v.append(i)

    # Print the Kth element
    print(v[K - 1])


# Driver Code
if __name__ == "__main__":
    N = 10
    K = 3
    findKthElement(N, K)

    # This code is contributed by ukasp.

C#

// C# program for above approach
using System;
using System.Collections.Generic;

public class GFG
{

  // Function to find the K-th element
  // in the required permutation
  static void findKthElement(int N, int K)
  {

    // Stores the required permutation
    List<int> v = new List<int>();

    // Insert all the even numbers
    // less than or equal to N
    for (int i = 1; i <= N; i++) {
      if (i % 2 == 0) {
        v.Add(i);
      }
    }

    // Now, insert all odd numbers
    // less than or equal to N
    for (int i = 1; i <= N; i++) {
      if (i % 2 != 0) {
        v.Add(i);
      }
    }

    // Print the Kth element
    Console.WriteLine(v[K - 1]);
  }

  // Driver code
  public static void Main(String[] args)
  {
    int N = 10, K = 3;

    // functions call
    findKthElement(N, K);
  }
}

// This code is contributed by susmitakundugoaldanga.

Javascript

<script>

// JavaScript program for the above approach

    // Function to find the K-th element
    // in the required permutation
    function findKthElement(N , K) {

        // Stores the required permutation
        var v = [];

        // Insert all the even numbers
        // less than or equal to N
        for (i = 1; i <= N; i++) {
            if (i % 2 == 0) {
                v.push(i);
            }
        }

        // Now, insert all odd numbers
        // less than or equal to N
        for (i = 1; i <= N; i++) {
            if (i % 2 != 0) {
                v.push(i);
            }
        }

        // Print the Kth element
        document.write(v[K - 1]);
    }

    // Driver code


        var N = 10, K = 3;

        // functions call
        findKthElement(N, K);

// This code contributed by Rajput-Ji

</script>

Output

6

Time Complexity: O(N)
Auxiliary Space: O(N)

Efficient Approach: To optimize the above approach, the idea is based on the observation that the first N / 2 elements are even and the value of the K^th element in the first half is equal to K * 2. If K > N/2, the value of the K^th element, depends on whether N is odd or even. 
Follow the steps below to solve the problem:

• Initialize a variable, say ans, to store the K^th element.
• Check if the value of K ≤ N/2. If found to be true, update ans to K*2.
• Otherwise, K lies in the second half. In this case, ans depends on the value of N.
  • If the value of N is even, update ans to (K*2)-N-1.
  • Otherwise, update ans to (K*2)-N.
• Print the value of ans as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the Kth element
// in the required permutation
void findKthElement(int N, int K)
{
    // Store the required result
    int ans = 0;

    // If K is in the first
    // N / 2 elements, print K * 2
    if (K <= N / 2) {
        ans = K * 2;
    }

    // Otherwise, K is greater than N/2
    else {

        // If N is even
        if (N % 2 == 0) {
            ans = (K * 2) - N - 1;
        }

        // If N is odd
        else {
            ans = (K * 2) - N;
        }
    }

    // Print the required result
    cout << ans;
}

// Driver Code
int main()
{
    int N = 10, K = 3;
    findKthElement(N, K);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {

  // Function to find the Kth element
  // in the required permutation
  static void findKthElement(int N, int K)
  {
    // Store the required result
    int ans = 0;

    // If K is in the first
    // N / 2 elements, print K * 2
    if (K <= N / 2) {
      ans = K * 2;
    }

    // Otherwise, K is greater than N/2
    else {

      // If N is even
      if (N % 2 == 0) {
        ans = (K * 2) - N - 1;
      }

      // If N is odd
      else {
        ans = (K * 2) - N;
      }
    }

    // Print the required result
    System.out.println(ans);
  }

  // Driver code
  public static void main(String[] args)
  {

    int N = 10, K = 3;

    // functions call
    findKthElement(N, K);
  }
}

// This code is contributed by Kingash.

Python3

# Python 3 program for the above approach

# Function to find the Kth element
# in the required permutation
def findKthElement(N, K):

    # Store the required result
    ans = 0

    # If K is in the first
    # N / 2 elements, print K * 2
    if (K <= N / 2):
        ans = K * 2

    # Otherwise, K is greater than N/2
    else:

        # If N is even
        if (N % 2 == 0):
            ans = (K * 2) - N - 1

        # If N is odd
        else:
            ans = (K * 2) - N

    # Print the required result
    print(ans)

# Driver Code
if __name__ == '__main__':
    N = 10
    K = 3
    findKthElement(N, K)

    # This code is contributed by ipg2016107.

C#

// C# program for the above approach
using System;

class GFG{

  // Function to find the Kth element
  // in the required permutation
  static void findKthElement(int N, int K)
  {

    // Store the required result
    int ans = 0;

    // If K is in the first
    // N / 2 elements, print K * 2
    if (K <= N / 2) {
      ans = K * 2;
    }

    // Otherwise, K is greater than N/2
    else {

      // If N is even
      if (N % 2 == 0) {
        ans = (K * 2) - N - 1;
      }

      // If N is odd
      else {
        ans = (K * 2) - N;
      }
    }

    // Print the required result
    Console.Write(ans);
  }

  // Driver code
  static void Main()
  {
    int N = 10, K = 3;

    // functions call
    findKthElement(N, K);
  }
}

// This code is contributed by sanjoy_62.

Javascript

<script>
// javascript program for the above approach

// Function to find the Kth element
// in the required permutation
function findKthElement( N,  K)
{

    // Store the required result
    let ans = 0;

    // If K is in the first
    // N / 2 elements, print K * 2
    if (K <= N / 2)
    {
        ans = K * 2;
    }

    // Otherwise, K is greater than N/2
    else
    {

        // If N is even
        if (N % 2 == 0)
        {
            ans = (K * 2) - N - 1;
        }

        // If N is odd
        else
        {
            ans = (K * 2) - N;
        }
    }

    // Print the required result
    document.write(ans);
}

// Driver Code
    let N = 10, K = 3;
    findKthElement(N, K);

// This code is contributed by todaysgaurav
</script>

Output

6

Time Complexity: O(1)
Auxiliary Space: O(1)