Find the Kth position element of the given sequence

Given two integers N and K, the task is to find the element at the Kth position if all odd numbers from 1 to N are written down in increasing order followed by all the even numbers from 1 to N in increasing order.

Examples:

Input: N = 10, K = 3
Output: 5
The required sequence is 1, 3, 5, 7, 9, 2, 4, 6, 8 and 10.



Input: N = 7, K = 7
Output: 6

Approach: It is known that the Nth even number is given by 2 * K and the Nth odd number is given by 2 * K – 1. But since the even numbers are written after (N + 1) / 2 odd numbers here. Therfore, Kth even number is given by 2 * (K – (N + 1) / 2) and the odd numbers will remain the same as 2 * K – 1

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 number
// from the required sequence
int kthNum(int n, int k)
{
  
    // Count of odd integers
    // in the sequence
    int a = (n + 1) / 2;
  
    // kth number is even
    if (k > a)
        return (2 * (k - a));
  
    // It is odd
    return (2 * k - 1);
}
  
// Driver code
int main()
{
    int n = 7, k = 7;
  
    cout << kthNum(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 number
// from the required sequence
static int kthNum(int n, int k)
{
  
    // Count of odd integers
    // in the sequence
    int a = (n + 1) / 2;
  
    // kth number is even
    if (k > a)
        return (2 * (k - a));
  
    // It is odd
    return (2 * k - 1);
}
  
// Driver code
public static void main(String []args) 
{
    int n = 7, k = 7;
  
    System.out.println(kthNum(n, k));
}
}
  
// This code is contributed by Rajput-Ji

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Python3

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# Python3 implementation of the approach 
  
# Function to return the kth number 
# from the required sequence 
def kthNum(n, k) :
  
    # Count of odd integers 
    # in the sequence 
    a = (n + 1) // 2
  
    # kth number is even 
    if (k > a) :
        return (2 * (k - a)); 
  
    # It is odd 
    return (2 * k - 1); 
  
# Driver code 
if __name__ == "__main__" :
  
    n = 7; k = 7
  
    print(kthNum(n, k)); 
  
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach
using System;
      
class GFG 
{
  
// Function to return the kth number
// from the required sequence
static int kthNum(int n, int k)
{
  
    // Count of odd integers
    // in the sequence
    int a = (n + 1) / 2;
  
    // kth number is even
    if (k > a)
        return (2 * (k - a));
  
    // It is odd
    return (2 * k - 1);
}
  
// Driver code
public static void Main(String []args) 
{
    int n = 7, k = 7;
  
    Console.WriteLine(kthNum(n, k));
}
}
  
// This code is contributed by PrinciRaj1992

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Output:

6

Time Complexity: O(1)



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