# Equally divide into two sets such that one set has maximum distinct elements

There are two processes P1 and P2, and N resources where N is an even number. There is an array of N size and arr[i] represents the type of ith resource.There may be more than one instance of a resource.You are to divide these resources equally between P1 and P2 such that maximum number of distinct number of resources are allocated to P2. Print maximum number of distinct resources allocated to P2.

Examples:

Input : arr[] = [1, 1, 2, 2, 3, 3]
Output:
Explanation:
There are three different kinds of resources (1, 2 and 3), and two for each kind. Optimal distribution: Process P1 has resources [1, 2, 3] and the process P2 has gifts [1, 2, 3], too. Process p2 has 3 distinct resources.

Input: arr[] = [1, 1, 2, 1, 3, 4]
Output:
Explanation:
There are three different kinds of resources (1, 2, 3, 4), 3 instances of 1 and  single instances of resource 2, 3, 4. Optimal distribution: Process P1 has resources [1, 1, 1] and the process P2 has gifts [2, 3, 4].
Process p2 has 3 distinct resources.

Approach 1 (Using sorting):

1. Sort the Array of resources.
2. Find out the elements which are unique by comparing the adjacent elements of the sorted array.suppose count holds the distinct number of resources in array.
3. Return the minimum of count and N/2.

Below is the implementation of the above approach:

## C++

 `// C++ program to equally divide n elements` `// into two sets such that second set has` `// maximum distinct elements.` `#include ` `#include ` `using` `namespace` `std;`   `int` `distribution(``int` `arr[], ``int` `n)` `{` `    ``sort(arr, arr + n);` `    ``int` `count = 1;` `    ``for` `(``int` `i = 1; i < n; i++) ` `        ``if` `(arr[i] > arr[i - 1])` `            ``count++;` `    `  `    ``return` `min(count, n / 2);` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 1, 1, 2, 1, 3, 4 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``cout << distribution(arr, n) << endl;` `    ``return` `0;` `}`

## Java

 `// Java program to equally divide n elements` `// into two sets such that second set has` `// maximum distinct elements.` `import` `java.util.*;` `class` `Geeks {` `    `  `static` `int` `distribution(``int` `arr[], ``int` `n)` `{` `    ``Arrays.sort(arr);` `    ``int` `count = ``1``;` `    ``for` `(``int` `i = ``1``; i < n; i++) ` `        ``if` `(arr[i] > arr[i - ``1``])` `            ``count++;` `    `  `    ``return` `Math.min(count, n / ``2``);` `}`   `// Driver code` `public` `static` `void` `main(String args[])` `{` `    ``int` `arr[] = { ``1``, ``1``, ``2``, ``1``, ``3``, ``4` `};` `    ``int` `n = arr.length;` `    ``System.out.println(distribution(arr, n));` `}` `}`   `// This code is contributed by ankita_saini`

## Python3

 `# Python 3 program to equally divide n ` `# elements into two sets such that second ` `# set has maximum distinct elements.`   `def` `distribution(arr, n):` `    ``arr.sort(reverse ``=` `False``)` `    ``count ``=` `1` `    ``for` `i ``in` `range``(``1``, n, ``1``):` `        ``if` `(arr[i] > arr[i ``-` `1``]):` `            ``count ``+``=` `1` `    `  `    ``return` `min``(count, n ``/` `2``)`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[``1``, ``1``, ``2``, ``1``, ``3``, ``4``] ` `    ``n ``=` `len``(arr)` `    ``print``(``int``(distribution(arr, n)))`   `# This code is contributed by` `# Shashank_Sharma`

## C#

 `// C# program to equally divide ` `// n elements into two sets such ` `// that second set has maximum` `// distinct elements.` `using` `System;`   `class` `GFG ` `{` `static` `int` `distribution(``int` `[]arr, ``int` `n)` `{` `    ``Array.Sort(arr);` `    ``int` `count = 1;` `    ``for` `(``int` `i = 1; i < n; i++) ` `        ``if` `(arr[i] > arr[i - 1])` `            ``count++;` `    `  `    ``return` `Math.Min(count, n / 2);` `}`   `// Driver code` `public` `static` `void` `Main(String []args)` `{` `    ``int` `[]arr= { 1, 1, 2, 1, 3, 4 };` `    ``int` `n = arr.Length;` `    ``Console.WriteLine(distribution(arr, n));` `}` `}`   `// This code is contributed ` `// by ankita_saini`

## PHP

 ` ``\$arr``[``\$i` `- 1])` `            ``\$count``++;` `    `  `    ``return` `min(``\$count``, ``\$n` `/ 2);` `}`   `// Driver code` `\$arr` `= ``array``(1, 1, 2, 1, 3, 4 );` `\$n` `= ``count``(``\$arr``);` `echo``(distribution(``\$arr``, ``\$n``));`   `// This code is contributed ` `// by inder_verma` `?>`

## Javascript

 ``

Output

`3`

Complexity Analysis:

• Time complexity: O(N log N)
• Auxiliary Space: O(1)

Approach 2(using hash set): Another way to find out distinct element is set, insert all the element in the set. By the property of a set, it will contain only unique elements. At the end, we can count the number of elements in the set, given by, say count. The value to be returned will again be given by min(count, n/2).

Implementation:

## C++

 `// C++ program to equally divide n elements` `// into two sets such that second set has` `// maximum distinct elements.` `#include ` `using` `namespace` `std;`   `int` `distribution(``int` `arr[], ``int` `n)` `{` `    ``set<``int``, greater<``int``> > resources;`   `    ``// Insert all the resources in the set` `    ``// There will be unique resources in the set` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``resources.insert(arr[i]);    `   `    ``// return minimum of distinct resources` `    ``// and n/2` `      ``int` `m = resources.size();` `    ``return` `min(m, n / 2);` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 1, 1, 2, 1, 3, 4 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``cout << distribution(arr, n) << endl;` `    ``return` `0;` `}`

## Java

 `// Java program to equally divide n elements` `// into two sets such that second set has` `// maximum distinct elements.` `import` `java.util.*;`   `class` `GFG` `{`   `static` `int` `distribution(``int` `arr[], ``int` `n)` `{` `    ``Set resources = ``new` `HashSet();`   `    ``// Insert all the resources in the set` `    ``// There will be unique resources in the set` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``resources.add(arr[i]); `   `    ``// return minimum of distinct resources` `    ``// and n/2` `    ``return` `Math.min(resources.size(), n / ``2``);` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `arr[] = { ``1``, ``1``, ``2``, ``1``, ``3``, ``4` `};` `    ``int` `n = arr.length;` `    ``System.out.print(distribution(arr, n) +``"\n"``);` `}` `}`   `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 program to equally divide n elements` `# into two sets such that second set has` `# maximum distinct elements.` `def` `distribution(arr, n):` `    ``resources ``=` `set``()` `    `  `    ``# Insert all the resources in the set` `    ``# There will be unique resources in the set` `    ``for` `i ``in` `range``(n):` `        ``resources.add(arr[i]);`   `        ``# return minimum of distinct resources` `        ``# and n/2` `    ``return` `min``(``len``(resources), n ``/``/` `2``);`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[ ``1``, ``1``, ``2``, ``1``, ``3``, ``4` `];` `    ``n ``=` `len``(arr);` `    ``print``(distribution(arr, n), "");` `    `  `# This code is contributed by PrinciRaj1992`

## C#

 `// C# program to equally divide n elements` `// into two sets such that second set has` `// maximum distinct elements.` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG` `{`   `static` `int` `distribution(``int` `[]arr, ``int` `n)` `{` `    ``HashSet<``int``> resources = ``new` `HashSet<``int``>();`   `    ``// Insert all the resources in the set` `    ``// There will be unique resources in the set` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``resources.Add(arr[i]); `   `    ``// return minimum of distinct resources` `    ``// and n/2` `    ``return` `Math.Min(resources.Count, n / 2);` `}`   `// Driver code` `public` `static` `void` `Main(String[] args)` `{` `    ``int` `[]arr = { 1, 1, 2, 1, 3, 4 };` `    ``int` `n = arr.Length;` `    ``Console.Write(distribution(arr, n) +``"\n"``);` `}` `}`   `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output

`3`

Complexity Analysis:

• Time Complexity: O(nlogn), where n is the size of the given array
• Auxiliary Space: O(n), as extra space of size n is used to create a set

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