前N个自然数置换的第K个元素，所有偶数均按升序排列在奇数之前

给定两个整数N和K ，任务是在排列的前N个自然数的排列中找到第K^个元素，以使所有偶数都按递增顺序出现在奇数之前。

例子 ：

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.

天真的方法：解决问题的最简单方法是生成所需的前N个自然数的排列，然后遍历该排列以找到其中存在的第K^个元素。
请按照以下步骤解决问题：

• 初始化一个数组，例如大小为N的V [] ，以存储所需的序列。
• 将所有小于或等于N的偶数插入V [] 。
• 然后，将所有小于或等于N的奇数插入V []中。
• 形成数组后，打印V [K – 1]的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the K-th element
// in the required permutation
void findKthElement(int N, int K)
{
    // Stores teh required permutation
    vector 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 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 teh 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 v = new List();
 
    // 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

 
// 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
 

输出

6

时间复杂度： O(N)
辅助空间： O(N)

高效方法：为了优化上述方法，该思想基于以下观察结果：前N / 2个元素为偶数，并且前一半中的第K^个元素的值等于K * 2 。如果K> N / 2 ，则第K^个元素的值取决于N是奇数还是偶数。
请按照以下步骤解决问题：

• 初始化一个变量，例如ans，以存储第K^个元素。
• 检查K的值是否≤N / 2 。如果确定为true，则将ans更新为K * 2 。
• 否则， K位于下半部分。在这种情况下， ans取决于N的值。
  • 如果N的值是偶数，则将ans更新为(K * 2)-N-1 。
  • 否则，将ans更新为(K * 2)-N 。
• 打印ans的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
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.

Java脚本

// 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

输出

6

时间复杂度： O(1)
辅助空间： O(1)