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

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: 3 
{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 <bits/stdc++.h>
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

<?php
// PHP implementation of the approach
// Function to return the maximized median
function getMaxMedian($arr, $n, $k)
{
    $size = $n + $k;

    // Sort the array
    sort($arr, $n);

    // If size is even
    if ($size % 2 == 0)
    {
        $median = (float)($arr[($size / 2) - 1] +
                          $arr[$size / 2]) / 2;
        return $median;
    }

    // If size is odd
    $median = $arr[$size / 2];
    return $median;
}

// Driver code
$arr = array( 3, 2, 3, 4, 2 );
$n = sizeof($arr);
$k = 2;
echo(getMaxMedian($arr, $n, $k));

// This code is Contributed by Code_Mech.

Javascript

<script>

// JavaScript implementation of the approach

// Function to return the maximized median
function getMaxMedian(arr, n, k) {
    let size = n + k;

    // Sort the array
    arr.sort((a, b) => a - b);

    // If size is even
    if (size % 2 == 0) {
        let median = (arr[(Math.floor(size / 2) - 1)] +
        arr[(Math.floor(size / 2))]) / 2;
        return median;
    }

    // If size is odd
    let median = arr[(Math.floor(size / 2))];
    return median;
}

// Driver code

let arr = [3, 2, 3, 4, 2];
let n = arr.length;
let k = 2;
document.write(getMaxMedian(arr, n, k));


// This code is contributed by _saurabh_jaiswal

</script>

Output

3

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