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

0 0 4 minutes read

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:

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

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!