Partition the array into two odd length groups with minimized absolute difference between their median

Given an array arr[] of positive integers of even length, the task is to partition these elements of arr[] into two groups each of odd length such that the absolute difference between the median of the two groups is minimized.
Examples:

Input: arr[] = { 1, 2, 3, 4, 5, 6 } 
Output: 1 
Explanation:
Group 1 can be [2, 4, 6] with median 4 
Group 2 can be [1, 3, 5] with median 3. 
The absolute difference between the two medians is 4 – 3 = 1, which can’t be minimized further with any kind of groupings formed.
Input: arr[] = { 15, 25, 35, 50 } 
Output: 10 
Explanation:
Group 1 can be [15, 25, 50] with median 25 
Group 2 can be [35] with median 35. 
The absolute difference between the two medians is 35 – 25 = 10, which can’t be minimized further with any kind of groupings formed. 

Approach:

• If the given array arr[] is sorted, the middle elements of arr[] will give the minimum difference.
• So divide the arr[] in such a way that these two elements will be the median of two new arrays of odd length.
• Therefore, put the n/2^th element of the arr[] in the first group and the (n/2 – 1)^th element of the arr[] in the second group as a median respectively.
• Then abs(arr[n/2] – arr[(n/2)-1]) is the minimum difference between the two new arrays.

Below is the implementation of the above approach: 

C++

// C++ program to minimize the
// median between partition array

#include "bits/stdc++.h"
using namespace std;

// Function to find minimize the
// median between partition array
int minimizeMedian(int arr[], int n)
{
    // Sort the given array arr[]
    sort(arr, arr + n);

    // Return the difference of two
    // middle element of the arr[]
    return abs(arr[n / 2] - arr[(n / 2) - 1]);
}

// Driver Code
int main()
{
    int arr[] = { 15, 25, 35, 50 };

    // Size of arr[]
    int n = sizeof(arr) / sizeof(arr[0]);

    // Function that returns the minimum
    // the absolute difference between
    // median of partition array
    cout << minimizeMedian(arr, n);
    return 0;
}

Java

// Java program to minimise the
// median between partition array
import java.util.*;

class GFG
{

    // Function to find minimise the
    // median between partition array
    static int minimiseMedian(int arr[], int n)
    {
        // Sort the given array arr[]
        Arrays.sort(arr);

        // Return the difference of two
        // middle element of the arr[]
        return Math.abs(arr[n / 2] - arr[(n / 2) - 1]);
    }

    // Driver Code
    public static void main (String[] args)
    {
        int arr[] = { 15, 25, 35, 50 };

        // Size of arr[]
        int n = arr.length;

        // Function that returns the minimum
        // the absolute difference between
        // median of partition array
        System.out.println(minimiseMedian(arr, n));
    }
}

// This code is contributed by AnkitRai01

Python3

# Python3 program to minimise the
# median between partition array

# Function to find minimise the
# median between partition array
def minimiseMedian(arr, n) :

    # Sort the given array arr[]
    arr.sort();

    # Return the difference of two
    # middle element of the arr[]
    ans = abs(arr[n // 2] - arr[(n // 2) - 1]);

    return ans;

# Driver Code
if __name__ == "__main__" :

    arr = [ 15, 25, 35, 50 ];

    # Size of arr[]
    n = len(arr);

    # Function that returns the minimum
    # the absolute difference between
    # median of partition array
    print(minimiseMedian(arr, n));

# This code is contributed by AnkitRai01

C#

// C# program to minimise the
// median between partition array
using System;

class GFG
{

    // Function to find minimise the
    // median between partition array
    static int minimiseMedian(int []arr, int n)
    {
        // Sort the given array []arr
        Array.Sort(arr);

        // Return the difference of two
        // middle element of the []arr
        return Math.Abs(arr[n / 2] - arr[(n / 2) - 1]);
    }

    // Driver Code
    public static void Main(String[] args)
    {
        int []arr = { 15, 25, 35, 50 };

        // Size of []arr
        int n = arr.Length;

        // Function that returns the minimum
        // the absolute difference between
        // median of partition array
        Console.WriteLine(minimiseMedian(arr, n));
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// JavaScript program to minimise the
// median between partition array

// Function to find minimise the
// median between partition array
function minimiseMedian(arr, n) {
    // Sort the given array arr[]
    arr.sort((a, b) => a - b);

    // Return the difference of two
    // middle element of the arr[]
    return Math.abs(arr[n / 2] - arr[(n / 2) - 1]);
}

// Driver Code
let arr = [15, 25, 35, 50];

// Size of arr[]
let n = arr.length;

// Function that returns the minimum
// the absolute difference between
// median of partition array
document.write(minimiseMedian(arr, n));

// This code is contributed by gfgking

</script>

Output:

10

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