k-th number in the Odd-Even sequence

Given two numbers n and k, find the k-th number in the Odd-Even sequence made of n. The Odd-Even sequence contains first contains all odd numbers from 1 to n then all even numbers in set 1 to n.

Examples :
Input :  n = 5, k = 3
Output : 5
In this example, the Odd-Even  is
{1, 3, 5, 2, 4}. 
The third number in sequence is 5.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach :

The first approach is simply make a Odd-Even sequence and then find k-th element in it.

C++

// CPP program to find k-th  
// element in the Odd-Even  
// sequence. 
#include <bits/stdc++.h> 
using namespace std; 
  
int findK(int n, int k) 
{ 
    vector<long> a; 
      
    // insert all the odd  
    // numbers from 1 to n.  
    for (int i = 1; i < n; i++)  
        if (i % 2 == 1) 
            a.push_back(i); 
      
    // insert all the even  
    // numbers from 1 to n. 
    for (int i = 1; i < n; i++)  
        if (i % 2 == 0) 
            a.push_back(i); 
      
    return (a[k - 1]); 
} 
  
// Driver code 
int main() 
{ 
    long n = 10, k = 3; 
    cout << findK(n, k) << endl; 
    return 0; 
} 

Java

// Java program to find k-th  
// element in the Odd-Even  
// sequence.  
import java.util.*; 
  
class GFG{ 
static int findK(int n, int k)  
{  
    ArrayList<Integer> a = new ArrayList<Integer>(n); 
      
    // insert all the odd  
    // numbers from 1 to n.  
    for (int i = 1; i < n; i++)  
        if (i % 2 == 1)  
            a.add(i);  
      
    // insert all the even  
    // numbers from 1 to n.  
    for (int i = 1; i < n; i++)  
        if (i % 2 == 0)  
            a.add(i);  
      
    return (a.get(k - 1));  
}  
  
// Driver code  
public static void main(String[] args)  
{  
    int n = 10, k = 3;  
    System.out.println(findK(n, k));  
}  
} 
// This code is contributed by mits 

Python3

# Python3 code to find  
# k-th element in the 
# Odd-Even sequence. 
  
def findK (n, k ): 
    a = list() 
      
    # insert all the odd  
    # numbers from 1 to n. 
    i = 1
    while i < n: 
        a.append(i) 
        i = i + 2
      
    # insert all the even  
    # numbers from 1 to n. 
    i = 2
    while i < n: 
        a.append(i) 
        i = i + 2
          
    return (a[k - 1]) 
  
# Driver code 
n = 10
k = 3
print(findK(n, k)) 
  
# This code is contributed  
# by "Sharad_Bhardwaj". 

C#

// C# program to find k-th  
// element in the Odd-Even  
// sequence.  
using System;  
using System.Collections;  
  
class GFG{ 
static int findK(int n, int k)  
{  
    ArrayList a = new ArrayList(n); 
      
    // insert all the odd  
    // numbers from 1 to n.  
    for (int i = 1; i < n; i++)  
        if (i % 2 == 1)  
            a.Add(i);  
      
    // insert all the even  
    // numbers from 1 to n.  
    for (int i = 1; i < n; i++)  
        if (i % 2 == 0)  
            a.Add(i);  
      
    return (int)(a[k - 1]);  
}  
  
// Driver code  
static void Main()  
{  
    int n = 10, k = 3;  
    Console.WriteLine(findK(n, k));  
}  
} 
// This code is contributed by mits 

PHP

<?php 
// PHP program to find k-th  
// element in the Odd-Even  
// sequence. 
  
function findK($n, $k) 
{ 
    $a; 
    $index = 0; 
      
    // insert all the odd  
    // numbers from 1 to n.  
    for ($i = 1; $i < $n; $i++)  
        if ($i % 2 == 1) 
            $a[$index++] = $i; 
      
    // insert all the even  
    // numbers from 1 to n. 
    for ($i = 1; $i < $n; $i++)  
        if ($i % 2 == 0) 
            $a[$index++] = $i; 
      
    return ($a[$k - 1]); 
} 
  
// Driver code 
$n = 10; 
$k = 3; 
echo findK($n, $k); 
  
// This code is contributed by mits. 
?> 

Output :

The time complexity of the above approach is O(n), for an O(1) approach read the approach discussed here.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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