等分为两组，使得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 N 大小的数组，arr[i] 表示第 i 个资源的类型。资源的实例可能不止一个。您要在 P1 和 P2 之间平均分配这些资源，以便最大数量的不同资源数量分配给P2。打印分配给 P2 的不同资源的最大数量。

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. 
 

方法一（使用排序）：

1. 对资源数组进行排序。
2. 通过比较已排序数组的相邻元素找出唯一的元素。假设 count 保存数组中不同数量的资源。
3. 返回计数和 N/2 的最小值。

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
?>

Javascript

 
// Javascript program to equally divide n elements
// into two sets such that second set has
// maximum distinct elements.
 
function distribution(arr, n)
{
    arr.sort((a,b)=>a-b);
    var count = 1;
    for (var i = 1; i < n; i++)
        if (arr[i] > arr[i - 1])
            count++;
     
    return Math.min(count, parseInt(n / 2));
}
 
// Driver code
var arr = [1, 1, 2, 1, 3, 4];
var n = arr.length;
document.write( distribution(arr, n));
 

输出：

3

时间复杂度– O(N log N)

方法二（使用散列集）
另一种找出不同元素的方法是集合，插入集合中的所有元素。根据集合的属性，它将只包含唯一元素。最后，我们可以计算集合中元素的数量，例如计数。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

蟒蛇3

# 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

Javascript

    // Javascript program to equally divide n elements
    // into two sets such that second set has
    // maximum distinct elements.
    function distribution(arr, n)
    {
        let resources = new Set();
 
        // Insert all the resources in the set
        // There will be unique resources in the set
        for (let i = 0; i < n; i++)
            resources.add(arr[i]);
 
        // return minimum of distinct resources
        // and n/2
        return Math.min(resources.size, parseInt(n / 2, 10));
    }
     
    let arr = [ 1, 1, 2, 1, 3, 4 ];
    let n = arr.length;
    document.write(distribution(arr, n) +"</br>");
 
// This code is contributed by mukesh07.

输出

3

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。