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

  • Difficulty Level : Easy
  • Last Updated : 27 Aug, 2021

Given two integers N and K, the task is to find the Kth 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 :

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

Input: N = 5, K = 4
Output: 3
Explanation:
The required permutation is {2, 4, 1, 3, 5}.
The 4th 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 Kth 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 Kth element in the first half is equal to K * 2. If K > N/2, the value of the Kth 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 Kth 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.
  • 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)




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