# Find Kth element in an array containing odd elements first and then even elements

Given the length of an array of integers N and an integer K. The task is to modify the array in such a way that the array contains first all odd integers from 1 to N in ascending order, then all even integers from 1 to N in ascending order and then print the Kth element in the modified array.

Examples:

Input: N = 8, K = 5
Output: 2
The array will be {1, 3, 5, 7, 2, 4, 6, 8}
and the fifth element is 2.

Input: N = 7, K = 2
Output: 3

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

Naive approach: A simple approach is to store the odd numbers first, one by one till N and then storing the even numbers one by one till N, and then printing the kth element.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the kth element ` `// in the modified array ` `int` `getNumber(``int` `n, ``int` `k) ` `{ ` `    ``int` `arr[n]; ` ` `  `    ``int` `i = 0; ` ` `  `    ``// First odd number ` `    ``int` `odd = 1; ` `    ``while` `(odd <= n) { ` ` `  `        ``// Insert the odd number ` `        ``arr[i++] = odd; ` ` `  `        ``// Next odd number ` `        ``odd += 2; ` `    ``} ` ` `  `    ``// First even number ` `    ``int` `even = 2; ` `    ``while` `(even <= n) { ` ` `  `        ``// Insert the even number ` `        ``arr[i++] = even; ` ` `  `        ``// Next even number ` `        ``even += 2; ` `    ``} ` ` `  `    ``// Return the kth element ` `    ``return` `arr[k - 1]; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 8, k = 5; ` `    ``cout << getNumber(n, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG  ` `{ ` ` `  `// Function to return the kth element ` `// in the modified array ` `static` `int` `getNumber(``int` `n, ``int` `k) ` `{ ` `    ``int` `[]arr = ``new` `int``[n]; ` ` `  `    ``int` `i = ``0``; ` ` `  `    ``// First odd number ` `    ``int` `odd = ``1``; ` `    ``while` `(odd <= n) ` `    ``{ ` ` `  `        ``// Insert the odd number ` `        ``arr[i++] = odd; ` ` `  `        ``// Next odd number ` `        ``odd += ``2``; ` `    ``} ` ` `  `    ``// First even number ` `    ``int` `even = ``2``; ` `    ``while` `(even <= n) ` `    ``{ ` ` `  `        ``// Insert the even number ` `        ``arr[i++] = even; ` ` `  `        ``// Next even number ` `        ``even += ``2``; ` `    ``} ` ` `  `    ``// Return the kth element ` `    ``return` `arr[k - ``1``]; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `n = ``8``, k = ``5``; ` `    ``System.out.println(getNumber(n, k)); ` `} ` `}  ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the kth element ` `# in the modified array ` `def` `getNumber(n, k): ` `    ``arr ``=` `[``0``] ``*` `n; ` ` `  `    ``i ``=` `0``; ` ` `  `    ``# First odd number ` `    ``odd ``=` `1``; ` `    ``while` `(odd <``=` `n): ` `         `  `        ``# Insert the odd number ` `        ``arr[i] ``=` `odd; ` `        ``i ``+``=` `1``; ` ` `  `        ``# Next odd number ` `        ``odd ``+``=` `2``; ` ` `  `    ``# First even number ` `    ``even ``=` `2``; ` `    ``while` `(even <``=` `n): ` `        ``# Insert the even number ` `        ``arr[i] ``=` `even; ` `        ``i ``+``=` `1``; ` ` `  `        ``# Next even number ` `        ``even ``+``=` `2``; ` ` `  `    ``# Return the kth element ` `    ``return` `arr[k ``-` `1``]; ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``n ``=` `8``; ` `    ``k ``=` `5``; ` `    ``print``(getNumber(n, k)); ` ` `  `# This code is contributed by Rajput-Ji `

## C#

 `// C# implementation of the approach ` `using` `System; ` `     `  `class` `GFG  ` `{ ` ` `  `// Function to return the kth element ` `// in the modified array ` `static` `int` `getNumber(``int` `n, ``int` `k) ` `{ ` `    ``int` `[]arr = ``new` `int``[n]; ` ` `  `    ``int` `i = 0; ` ` `  `    ``// First odd number ` `    ``int` `odd = 1; ` `    ``while` `(odd <= n) ` `    ``{ ` ` `  `        ``// Insert the odd number ` `        ``arr[i++] = odd; ` ` `  `        ``// Next odd number ` `        ``odd += 2; ` `    ``} ` ` `  `    ``// First even number ` `    ``int` `even = 2; ` `    ``while` `(even <= n) ` `    ``{ ` ` `  `        ``// Insert the even number ` `        ``arr[i++] = even; ` ` `  `        ``// Next even number ` `        ``even += 2; ` `    ``} ` ` `  `    ``// Return the kth element ` `    ``return` `arr[k - 1]; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `n = 8, k = 5; ` `    ``Console.WriteLine(getNumber(n, k)); ` `} ` `}  ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```2
```

Efficient approach: Find the index where the first even element will be stored in the generated array. Now if the value of k is less then or equal to index then the desired number will be k * 2 – 1 else the desired number will be (k – index) * 2

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the kth element ` `// in the modified array ` `int` `getNumber(``int` `n, ``int` `k) ` `{ ` `    ``int` `pos; ` ` `  `    ``// Finding the index from where the ` `    ``// even numbers will be stored ` `    ``if` `(n % 2 == 0) { ` `        ``pos = n / 2; ` `    ``} ` `    ``else` `{ ` `        ``pos = (n / 2) + 1; ` `    ``} ` ` `  `    ``// Return the kth element ` `    ``if` `(k <= pos) { ` `        ``return` `(k * 2 - 1); ` `    ``} ` `    ``else` ` `  `        ``return` `((k - pos) * 2); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 8, k = 5; ` ` `  `    ``cout << getNumber(n, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.io.*; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to return the kth element ` `// in the modified array ` `static` `int` `getNumber(``int` `n, ``int` `k) ` `{ ` `    ``int` `pos; ` ` `  `    ``// Finding the index from where the ` `    ``// even numbers will be stored ` `    ``if` `((n % ``2``) == ``0``)  ` `    ``{ ` `        ``pos = n / ``2``; ` `    ``} ` `    ``else`  `    ``{ ` `        ``pos = (n / ``2``) + ``1``; ` `    ``} ` ` `  `    ``// Return the kth element ` `    ``if` `(k <= pos) ` `    ``{ ` `        ``return` `(k * ``2` `- ``1``); ` `    ``} ` `    ``else` `        ``return` `((k - pos) * ``2``); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main (String[] args) ` `{ ` `    ``int` `n = ``8``, k = ``5``; ` `    ``System.out.println (getNumber(n, k)); ` `} ` `} ` ` `  `// This code is contributed by @tushil.  `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the kth element  ` `# in the modified array  ` `def` `getNumber(n, k) :  ` ` `  `    ``# Finding the index from where the  ` `    ``# even numbers will be stored  ` `    ``if` `(n ``%` `2` `=``=` `0``) :  ` `        ``pos ``=` `n ``/``/` `2``;  ` `     `  `    ``else` `: ` `        ``pos ``=` `(n ``/``/` `2``) ``+` `1``;  ` ` `  `    ``# Return the kth element  ` `    ``if` `(k <``=` `pos) : ` `        ``return` `(k ``*` `2` `-` `1``);  ` `         `  `    ``else` `: ` `        ``return` `((k ``-` `pos) ``*` `2``);  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `: ` `    ``n ``=` `8``; k ``=` `5``; ` `     `  `    ``print``(getNumber(n, k));  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `         `  `// Function to return the kth element ` `// in the modified array ` `static` `int` `getNumber(``int` `n, ``int` `k) ` `{ ` `    ``int` `pos; ` ` `  `    ``// Finding the index from where the ` `    ``// even numbers will be stored ` `    ``if` `((n % 2) == 0)  ` `    ``{ ` `        ``pos = n / 2; ` `    ``} ` `    ``else` `    ``{ ` `        ``pos = (n / 2) + 1; ` `    ``} ` ` `  `    ``// Return the kth element ` `    ``if` `(k <= pos) ` `    ``{ ` `        ``return` `(k * 2 - 1); ` `    ``} ` `    ``else` `        ``return` `((k - pos) * 2); ` `} ` ` `  `// Driver code ` `static` `public` `void` `Main () ` `{ ` `    ``int` `n = 8, k = 5; ` `    ``Console.Write(getNumber(n, k)); ` `} ` `} ` ` `  `// This code is contributed by @ajit. `

Output:

```2
```

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