# 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:1Explanation:

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:10Explanation:

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**element of the arr[] in the first group and the^{th}**(n/2 â€“ 1)**element of the^{th}**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)