由给定 N 个整数形成的所有对的成对差的平均值

给定一个包含N个整数的数组arr[] ，任务是计算由给定N个整数形成的两个元素之间的差的平均值。

例子：

Input: arr[] = {-1, 3, -5, 4}
Output: 5.166667
Explanation: There are 6 possible pair of points in the given array with the pairwise difference as: diff(-1, 3) = 4, diff(-1, -5) = 4, diff(-1, 4) = 5, diff(3, -5) = 8, diff(3, 4) = 1, diff(-5, 4) = 9. Therefore, average pairwise difference is (4 + 4 + 5 + 8 + 1 + 9)/6 = 31/6 = 5.166667.

Input: arr[] = { -1, 2, -3, 7, -6 }
Output: 6.2

方法：这个问题可以通过使用贪心方法和前缀和方法来解决。如果数组arr[]中的点按排序顺序排列，则第 i^个点到所有较大点的距离之和可以计算为： (arr[i+1] – arr[i]) + (arr[ i+2] – arr[i]) … + (arr[N-1] – arr[i]) => (arr[i+1] + arr[i+2]… + arr[N-1]) – arr[i] * (N – 1 – i) 。使用此观察，可以使用以下步骤解决给定问题：

• 最初以非递减顺序对数组arr[]进行排序。
• 创建数组arr[]的前缀和数组pre[ ] 。
• 遍历每个索引i并将(pre[N – 1] – pre[i]) – arr[i] * (N – 1 – i)添加到变量ans中。
• 所需的答案是ans / count of pairs => ans / (N*(N-1)/2) 。

下面是上述方法的实现：

C++

// C++ program for above approach
#include 
using namespace std;
 
// Function to find average distance
// between given points on a line
long double averageDistance(
  vector arr, int N)
{
    // Sorting the array arr[]
    sort(arr.begin(), arr.end());
 
    // Stores the prefix sum
    // array of arr[]
    int pre[N] = { 0 };
    pre[0] = arr[0];
 
    // Loop to calculate prefix sum
    for (int i = 1; i < N; i++) {
        pre[i] = pre[i - 1] + arr[i];
    }
 
    // Initialising the answer variable
    long double ans = 0;
 
    // Loop to iterate through arr[]
    for (int i = 0; i < N - 1; i++) {
 
        // Adding summation of all
        // distances from ith point
        ans += (pre[N - 1] - pre[i])
          - arr[i] * (N - 1 - i);
    }
 
    // Return Average
    return ans / ((N * (N - 1)) / 2);
}
 
// Driver Code
int main()
{
    vector arr = { -1, 3, -5, 4 };
    cout << averageDistance(arr, arr.size());
 
    return 0;
}

Java

// Java implementation for the above approach
import java.util.*;
class GFG {
 
    // Function to find average distance
    // between given points on a line
    static double averageDistance(int[] arr, int N)
    {
        // Sorting the array arr[]
        Arrays.sort(arr);
 
        // Stores the prefix sum
        // array of arr[]
        int[] pre = new int[N];
        pre[0] = arr[0];
 
        // Loop to calculate prefix sum
        for (int i = 1; i < N; i++) {
            pre[i] = pre[i - 1] + arr[i];
        }
 
        // Initialising the answer variable
        double ans = 0;
 
        // Loop to iterate through arr[]
        for (int i = 0; i < N - 1; i++) {
 
            // Adding summation of all
            // distances from ith point
            ans += (pre[N - 1] - pre[i])
                   - arr[i] * (N - 1 - i);
        }
 
        // Return Average
        ans = (ans / ((N * (N - 1)) / 2));
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { -1, 3, -5, 4 };
 
        System.out.print(String.format(
            "%.5f", averageDistance(arr, arr.length)));
    }
}
 
// This code is contributed by ukasp.

Python3

# Python3 program for above approach
 
# Function to find average distance
# between given points on a line
def averageDistance(arr, N):
     
    # Sorting the array arr[]
    arr.sort()
 
    # Stores the prefix sum
    # array of arr[]
    pre = [0 for _ in range(N)]
    pre[0] = arr[0]
 
    # Loop to calculate prefix sum
    for i in range(1, N):
        pre[i] = pre[i - 1] + arr[i]
 
    # Initialising the answer variable
    ans = 0
 
    # Loop to iterate through arr[]
    for i in range(0, N - 1):
         
        # Adding summation of all
        # distances from ith point
        ans += ((pre[N - 1] - pre[i]) -
               (arr[i] * (N - 1 - i)))
 
    # Return Average
    return ans / ((N * (N - 1)) / 2)
 
# Driver Code
if __name__ == "__main__":
 
    arr = [ -1, 3, -5, 4 ]
     
    print(averageDistance(arr, len(arr)))
 
# This code is contributed by rakeshsahni

C#

// C# implementation for the above approach
using System;
class GFG
{
 
// Function to find average distance
// between given points on a line
static double averageDistance(
  int []arr, int N)
{
    // Sorting the array arr[]
    Array.Sort(arr);
 
    // Stores the prefix sum
    // array of arr[]
    int []pre = new int[N];
    pre[0] = arr[0];
 
    // Loop to calculate prefix sum
    for (int i = 1; i < N; i++) {
        pre[i] = pre[i - 1] + arr[i];
    }
 
    // Initialising the answer variable
    double ans = 0;
 
    // Loop to iterate through arr[]
    for (int i = 0; i < N - 1; i++) {
 
        // Adding summation of all
        // distances from ith point
        ans += (pre[N - 1] - pre[i])
          - arr[i] * (N - 1 - i);
    }
 
    // Return Average
    ans  = Math.Round((ans / ((N * (N - 1)) / 2)), 5);
    return ans;
}
 
// Driver Code
public static void Main()
{
    int []arr = { -1, 3, -5, 4 };
     
    Console.Write(averageDistance(arr, arr.Length));
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

 
      // JavaScript Program to implement
      // the above approach
 
      // Function to find average distance
      // between given points on a line
      function averageDistance(
          arr, N)
      {
       
          // Sorting the array arr[]
          arr.sort(function (a, b) { return a - b })
 
          // Stores the prefix sum
          // array of arr[]
          let pre = new Array(N).fill(0);
          pre[0] = arr[0];
 
          // Loop to calculate prefix sum
          for (let i = 1; i < N; i++) {
              pre[i] = pre[i - 1] + arr[i];
          }
 
          // Initialising the answer variable
          let ans = 0;
 
          // Loop to iterate through arr[]
          for (let i = 0; i < N - 1; i++) {
 
              // Adding summation of all
              // distances from ith point
              ans += (pre[N - 1] - pre[i])
                  - arr[i] * (N - 1 - i);
          }
 
          // Return Average
          return ans / ((N * (N - 1)) / 2);
      }
 
      // Driver Code
      let arr = [-1, 3, -5, 4];
      document.write(averageDistance(arr, arr.length).toPrecision(6));
 
  // This code is contributed by Potta Lokesh
  

输出

5.16667

时间复杂度： O(N*log N)
辅助空间： O(1)