m 个元素的两个子集之间的最大差异

给定一个由 n 个整数和一个数字 m 组成的数组，找出从给定数组中选择的两组 m 个元素之间的最大可能差异。
例子：

Input : arr[] = 1 2 3 4 5
            m = 4
Output : 4
The maximum four elements are 2, 3, 
4 and 5. The minimum four elements are 
1, 2, 3 and 4. The difference between
two sums is (2 + 3 + 4 + 5) - (1 + 2
+ 3 + 4) = 4
  
Input : arr[] = 5 8 11 40 15
           m = 2
Output : 42
The difference is (40 + 15) - (5  + 8)           

这个想法是首先对数组进行排序，然后找到前 m 个元素的总和和最后 m 个元素的总和。最后返回两个和之间的差值。

CPP

// C++ program to find difference
// between max and min sum of array
#include 
using namespace std;
 
// utility function
int find_difference(int arr[], int n, int m)
{
    int max = 0, min = 0;
 
    // sort array
    sort(arr, arr + n);
 
    for (int i = 0, j = n - 1;
         i < m; i++, j--) {
        min += arr[i];
        max += arr[j];
    }
 
    return (max - min);
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int m = 4;
    cout << find_difference(arr, n, m);
    return 0;
}

Java

// Java program to find difference
// between max and min sum of array
import java.util.Arrays;
 
class GFG {
    // utility function
    static int find_difference(int arr[], int n,
                               int m)
    {
        int max = 0, min = 0;
 
        // sort array
        Arrays.sort(arr);
 
        for (int i = 0, j = n - 1;
             i < m; i++, j--) {
            min += arr[i];
            max += arr[j];
        }
 
        return (max - min);
    }
 
    // Driver program
    public static void main(String arg[])
    {
        int arr[] = { 1, 2, 3, 4, 5 };
        int n = arr.length;
        int m = 4;
        System.out.print(find_difference(arr, n, m));
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python program to
# find difference
# between max and
# min sum of array
 
def find_difference(arr, n, m):
    max = 0; min = 0
      
    # sort array
    arr.sort();
    j = n-1
    for i in range(m):
        min += arr[i]
        max += arr[j]
        j = j - 1
      
    return (max - min)
  
# Driver code
if __name__ == "__main__":
    arr = [1, 2, 3, 4, 5]
    n = len(arr)
    m = 4
 
    print(find_difference(arr, n, m))  
 
# This code is contributed by
# Harshit Saini

C#

// C# program to find difference
// between max and min sum of array
using System;
 
class GFG {
     
    // utility function
    static int find_difference(int[] arr, int n,
                                          int m)
    {
        int max = 0, min = 0;
 
        // sort array
        Array.Sort(arr);
 
        for (int i = 0, j = n - 1;
            i < m; i++, j--) {
            min += arr[i];
            max += arr[j];
        }
 
        return (max - min);
    }
 
    // Driver program
    public static void Main()
    {
        int[] arr = { 1, 2, 3, 4, 5 };
        int n = arr.Length;
        int m = 4;
        Console.Write(find_difference(arr, n, m));
    }
}
 
// This code is contributed by nitin mittal

Javascript

// JavaScript program to find difference
// between max and min sum of array
 
// utility function
    function find_difference(arr, n,
                               m)
    {
        let max = 0, min = 0;
   
        // sort array
        arr.sort();
   
        for (let i = 0, j = n - 1;
             i < m; i++, j--) {
            min += arr[i];
            max += arr[j];
        }
   
        return (max - min);
    }
 
// Driver Code
 
        let arr = [ 1, 2, 3, 4, 5 ];
        let n = arr.length;
        let m = 4;
        document.write(find_difference(arr, n, m));
         
        // This code is contributed by splevel62.

输出：

4

我们可以使用下面文章中讨论的更有效的方法来优化上述解决方案。
数组中的 k 个最大（或最小）元素 |添加了最小堆方法

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