将数组拆分为最小数量的子集，以使所有对中的元素至少出现一次在不同的子集中

给定由N个不同的整数组成的数组arr [] ，任务是查找需要将数组拆分为两个子集的最小次数，以使每对元素至少出现一次为两个不同的子集。

例子：

Input: arr[] = { 3, 4, 2, 1, 5 } 
Output: 3 
Explanation: 
Possible pairs are { (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5) } 
Splitting the array into { 1, 2 } and { 3, 4, 5 } 
Elements of each of the pairs { (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5) } are present into two different subsets. 
Splitting the array into { 1, 3 } and { 2, 4, 5 } 
Elements of each of the pairs { (1, 2), (1, 4), (1, 5), (2, 3), (3, 4), (3, 5) } are present into two different subsets. 
Splitting the array into { 1, 3, 4 } and { 2, 5 } 
Elements of each of the pairs { (1, 2), (1, 5), (2, 3), (3, 5), (2, 4), (4, 5) } are present into two different subsets. 
Since elements of each pair of the array is present in two different subsets at least once, the required output is 3.

Input: arr[] = { 2, 1, 3 } 
Output: 2 
 

方法：想法是总是将数组分成大小为floor(N / 2)和ceil(N / 2)的两个子集。在每个分区之前，只需将arr [i]的值与arr [N / 2 + i]交换即可。请按照以下步骤解决问题：

• 如果N为奇数，则总可能的swap(arr [i]，arr [N / 2 + i])等于N / 2 + 1 。因此，打印(N / 2 + 1) 。
• 如果N为偶数，则总可能的swap(arr [i]，arr [N / 2 + i])等于N / 2 。因此，请打印(N / 2) 。

以下是上述方法的C++实现：

C++

// C++ program to to implement
// the above approach
#include 
using namespace std;
 
// Function to find minimum count of ways to split
// the array into two subset such that elements of
// each pair occurs in two different subset
int MinimumNoOfWays(int arr[], int n)
{
 
    // Stores minimum count of ways to split array
    // into two subset such that elements of
    // each pair occurs in two different subset
    int mini_no_of_ways;
 
    // If N is odd
    if (n % 2 == 0) {
        mini_no_of_ways = n / 2;
    }
    else {
        mini_no_of_ways = n / 2 + 1;
    }
    return mini_no_of_ways;
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 4, 2, 1, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << MinimumNoOfWays(arr, N);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
   
class GFG{
   
// Function to find minimum count of ways to split
// the array into two subset such that elements of
// each pair occurs in two different subset
static int MinimumNoOfWays(int arr[], int n)
{
  
    // Stores minimum count of ways to split array
    // into two subset such that elements of
    // each pair occurs in two different subset
    int mini_no_of_ways;
  
    // If N is odd
    if (n % 2 == 0) {
        mini_no_of_ways = n / 2;
    }
    else {
        mini_no_of_ways = n / 2 + 1;
    }
    return mini_no_of_ways;
}
   
// Driver code
public static void main(String[] args)
{
    int arr[] = { 3, 4, 2, 1, 5 };
    int N = arr.length;
    System.out.print(MinimumNoOfWays(arr, N));
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python program to to implement
# the above approach
 
# Function to find minimum count of ways to split
# the array into two subset such that elements of
# each pair occurs in two different subset
def MinimumNoOfWays(arr, n):
   
    # Stores minimum count of ways to split array
    # into two subset such that elements of
    # each pair occurs in two different subset
    min_no_of_ways = 0
     
    # if n is even
    if (n % 2 == 0):
        mini_no_of_ways = n // 2
         
    # n is odd
    else:
        mini_no_of_ways = n // 2 + 1
         
    return mini_no_of_ways
 
# driver code
if __name__ == '__main__':
    arr = [3, 4, 1, 2, 5]
    n = len(arr)
    print(MinimumNoOfWays(arr, n))
 
# This code is contributed by MuskanKalra1

C#

// C# program to implement
// the above approach
using System;
class GFG
{
   
// Function to find minimum count of ways to split
// the array into two subset such that elements of
// each pair occurs in two different subset
static int MinimumNoOfWays(int []arr, int n)
{
  
    // Stores minimum count of ways to split array
    // into two subset such that elements of
    // each pair occurs in two different subset
    int mini_no_of_ways;
  
    // If N is odd
    if (n % 2 == 0)
    {
        mini_no_of_ways = n / 2;
    }
    else
    {
        mini_no_of_ways = n / 2 + 1;
    }
    return mini_no_of_ways;
}
   
  // Driver code
  public static void Main(string[] args)
  {
      int[] arr = { 3, 4, 2, 1, 5 };
      int N = arr.Length;
      Console.WriteLine(MinimumNoOfWays(arr, N));
  }
}
 
// This code is contributed by AnkThon

Javascript

// javascript program to implement
// the above approach
 
    // Function to find minimum count of ways to split
    // the array into two subset such that elements of
    // each pair occurs in two different subset
    function MinimumNoOfWays(arr , n) {
 
        // Stores minimum count of ways to split array
        // into two subset such that elements of
        // each pair occurs in two different subset
        var mini_no_of_ways;
 
        // If N is odd
        if (n % 2 == 0) {
            mini_no_of_ways = n / 2;
        } else {
            mini_no_of_ways = n / 2 + 1;
        }
        return parseInt(mini_no_of_ways);
    }
 
    // Driver code
     
        var arr = [ 3, 4, 2, 1, 5 ];
        var N = arr.length;
        document.write(MinimumNoOfWays(arr, N));
 
// This code contributed by gauravrajput1

输出：

3

时间复杂度： O(1)
空间复杂度： O(1)