平均分为两组，以便一组具有最大不同的元素

有两个进程P1和P2，以及N个资源，其中N是偶数。有一个大小为N的数组，并且arr [i]表示第i个资源的类型。一个资源的实例可能不止一个。要在P1和P2之间平均分配这些资源，以使最大数量的不同资源数分配给P2。打印分配给P2的最大不同资源数量。

例子：

Input : arr[] = [1, 1, 2, 2, 3, 3] 
Output: 3 
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: 3 
Explanation: 
There are three different kinds of resources (1, 2, 3, 4), 3 instances of 1 and single 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. 
 

方法1(使用排序)：

1. 对资源数组进行排序。
2. 通过比较排序后的数组的相邻元素找出唯一的元素。假设count在数组中拥有不同数量的资源。
3. 返回最小计数和N / 2。

下面是上述方法的实现：

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
?>

输出：

3

时间复杂度– O(N log N)

方法2(使用哈希集)
找出不同元素的另一种方法是设置，将所有元素插入到集合中。通过集合的属性，它将仅包含唯一元素。最后，我们可以计算集合中元素的数量，例如count。要返回的值将再次由min(count，n / 2)给出。

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 > 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 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.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

输出

3