# Maximize the median of the given array after adding K elements to the same array

Last Updated : 24 Mar, 2023

Given an array arr[] of N elements and an integer K where K < N. The task is to insert K integer elements to the same array such that the median of the resultant array is maximized. Print the maximized median.
Examples:

Input: arr[] = {3, 2, 3, 4, 2}, k = 2
Output:
{2, 2, 3, 3, 4, 5, 5} can be once such resultant array with 3 as the median.
Input: arr[] = {3, 2, 3, 4, 2}, k = 3
Output: 3.5

Approach: In order to maximize the median of the resultant array, all the elements that need to be inserted must be greater than the maximum element from the array. After inserting these elements, the new size of the array will be size = N + K. Sort the array and the median of the array will be arr[size / 2] if the size is odd else (arr[(size / 2) – 1] + arr[size / 2]) / 2.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the maximized median ` `float` `getMaxMedian(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``int` `size = n + k; ` ` `  `    ``// Sort the array ` `    ``sort(arr, arr + n); ` ` `  `    ``// If size is even ` `    ``if` `(size % 2 == 0) { ` `        ``float` `median = (``float``)(arr[(size / 2) - 1] ` `                               ``+ arr[size / 2]) ` `                       ``/ 2; ` `        ``return` `median; ` `    ``} ` ` `  `    ``// If size is odd ` `    ``float` `median = arr[size / 2]; ` `    ``return` `median; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 3, 2, 3, 4, 2 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``int` `k = 2; ` `    ``cout << getMaxMedian(arr, n, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `import` `java.util.*; ` ` `  `// Java implementation of the approach ` `class` `GFG  ` `{ ` ` `  `    ``// Function to return the maximized median ` `    ``static` `double` `getMaxMedian(``int``[] arr, ``int` `n, ``int` `k)  ` `    ``{ ` `        ``int` `size = n + k; ` ` `  `        ``// Sort the array ` `        ``Arrays.sort(arr); ` ` `  `        ``// If size is even ` `        ``if` `(size % ``2` `== ``0``) ` `        ``{ ` `            ``double` `median = (``double``) (arr[(size / ``2``) - ``1``] ` `                    ``+ arr[size / ``2``]) ` `                    ``/ ``2``; ` `            ``return` `median; ` `        ``} ` ` `  `        ``// If size is odd ` `        ``double` `median1 = arr[size / ``2``]; ` `        ``return` `median1; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int``[] arr = {``3``, ``2``, ``3``, ``4``, ``2``}; ` `        ``int` `n = arr.length; ` `        ``int` `k = ``2``; ` `        ``System.out.print((``int``)getMaxMedian(arr, n, k)); ` ` `  `    ``} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Python 3 implementation of the approach ` ` `  `# Function to return the maximized median ` `def` `getMaxMedian(arr, n, k): ` `    ``size ``=` `n ``+` `k ` ` `  `    ``# Sort the array ` `    ``arr.sort(reverse ``=` `False``) ` ` `  `    ``# If size is even ` `    ``if` `(size ``%` `2` `=``=` `0``): ` `        ``median ``=` `(arr[``int``(size ``/` `2``) ``-` `1``] ``+`  `                  ``arr[``int``(size ``/` `2``)]) ``/` `2` `        ``return` `median ` ` `  `    ``# If size is odd ` `    ``median ``=` `arr[``int``(size ``/` `2``)] ` `    ``return` `median ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[``3``, ``2``, ``3``, ``4``, ``2``] ` `    ``n ``=` `len``(arr) ` `    ``k ``=` `2` `    ``print``(getMaxMedian(arr, n, k)) ` ` `  `# This code is contributed by ` `# Surendra_Gangwar `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Linq; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to return the maximized median ` `static` `double` `getMaxMedian(``int` `[]arr, ``int` `n, ``int` `k) ` `{ ` `    ``int` `size = n + k; ` ` `  `    ``// Sort the array ` `    ``Array.Sort(arr); ` ` `  `    ``// If size is even ` `    ``if` `(size % 2 == 0) ` `    ``{ ` `        ``double` `median = (``double``)(arr[(size / 2) - 1] ` `                            ``+ arr[size / 2]) ` `                    ``/ 2; ` `        ``return` `median; ` `    ``} ` ` `  `    ``// If size is odd ` `    ``double` `median1 = arr[size / 2]; ` `    ``return` `median1; ` `} ` ` `  `// Driver code ` `static` `void` `Main() ` `{ ` `    ``int` `[]arr = { 3, 2, 3, 4, 2 }; ` `    ``int` `n = arr.Length; ` `    ``int` `k = 2; ` `    ``Console.WriteLine(getMaxMedian(arr, n, k)); ` `} ` `} ` ` `  `// This code is contributed by mits `

## PHP

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## Javascript

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Output

`3`

Time Complexity: O(N * logN)
Auxiliary Space: O(1)