最大化arr [i] * i的总和

给定一个由N个整数组成的数组。您可以重新排列数组的元素。任务是找到Σarr[i] * i的最大值，其中i = 0、1、2，…。，n – 1。

例子：

Input : N = 4, arr[] = { 3, 5, 6, 1 }
Output : 31
If we arrange arr[] as { 1, 3, 5, 6 }. 
Sum of arr[i]*i is 1*0 + 3*1 + 5*2 + 6*3 
= 31, which is maximum

Input : N = 2, arr[] = { 19, 20 }
Output : 20

一个简单的解决方案是生成给定数组的所有排列。对于每个排列，计算Σarr[i] * i的值，最后返回最大值。

一种有效的解决方案基于以下事实：应将最大值缩放为最大值，而将最小值缩放为最小值。因此，我们将i的最小值乘以arr [i]的最小值。因此，以递增的顺序对给定的数组进行排序，并计算ari * i的总和，其中i = 0至n-1。

以下是此方法的实现：

C++

// CPP program to find the maximum value
// of i*arr[i]
#include
using namespace std;
  
int maxSum(int arr[], int n)
{  
  // Sort the array
  sort(arr, arr + n);
  
  // Finding the sum of arr[i]*i
  int sum = 0;
  for (int i = 0; i < n; i++)
    sum += (arr[i]*i);
  
  return sum;
}
  
// Driven Program
int main()
{
  int arr[] = { 3, 5, 6, 1 };
  int n = sizeof(arr)/sizeof(arr[0]);
  
  cout << maxSum(arr, n) << endl;
  return 0;
}

Java

// Java program to find the
// maximum value of i*arr[i]
import java.util.*;
  
class GFG {
  
    static int maxSum(int arr[], int n)
    {    
    // Sort the array
    Arrays.sort(arr);
  
    // Finding the sum of arr[i]*i
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += (arr[i] * i);
  
    return sum;
    }
  
    // Driven Program
    public static void main(String[] args)
    {
    int arr[] = { 3, 5, 6, 1 };
    int n = arr.length;
  
    System.out.println(maxSum(arr, n));
  
    }
}
// This code is contributed by Prerna Saini

Python3

# Python program to find the
# maximum value of i*arr[i]
def maxSum(arr,n):
  
    #  Sort the array
    arr.sort()
  
    # Finding the sum of 
    # arr[i]*i
    sum = 0
    for i in range(n):
        sum += arr[i] * i
          
    return sum
  
# Driver Program
arr = [3,5,6,1]
n = len(arr)
print(maxSum(arr,n))
  
# This code is contributed
# by Shrikant13

C#

// C# program to find the
// maximum value of i*arr[i]
using System;
  
class GFG {
      
    // Function to find the
    // maximum value of i*arr[i]
    static int maxSum(int[] arr, int n)
    { 
          
        // Sort the array
        Array.Sort(arr);
      
        // Finding the sum of arr[i]*i
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += (arr[i] * i);
      
        return sum;
    }
  
    // Driver code
    static public void Main()
    {
        int[] arr = {3, 5, 6, 1};
        int n = arr.Length;
      
        Console.WriteLine(maxSum(arr, n));
  
    }
}
  
// This code is contributed by Ajit.

输出：

31

时间复杂度：O(n Log n)