与数组中元素的最小差异总和

给定一个数组，我们需要在更改元素后找到数组元素的总和，因为arr [i]将变为abs(arr [i] -x)，其中x是数组元素。

例子：

Input  : {2, 5, 1, 7, 4}
Output : 9
We get minimum sum when we choose
x = 4. The minimum sum is 
abs(2-4) + abs(5-4) + abs(1-4) + (7-4)
abs(4-4) = 9

Input  : {5, 11, 14, 10, 17, 15}
Output : 20
We can either choose x = 14 or x = 11

这个想法是基于这样一个事实，即中间元素会导致差异的总和最小。当元素数为偶数时，我们可以取中间两个元素中的任何一个。我们可以通过几个例子来验证这一事实。

下面是上述想法的实现：

C++

// C++ program to find minimum sum of absolute
// differences with an array element.
#include
using namespace std;
    // function to find min sum after operation
    int absSumDidd(int a[],int n)
    {
        // Sort the array
        sort(a,a+n);
          
        // Pick middle value
        int midValue = a[(int)(n / 2)];
  
        // Sum of absolute differences.
        int sum = 0;
        for (int i = 0; i < n; i++) {
            sum = sum + abs(a[i] - midValue);
        }
          
        return sum;     
    }
  
    // Driver Code
    int main()
    {
        int arr[] = { 5, 11, 14, 10, 17, 15 };
        int n=sizeof(arr)/sizeof(arr[0]);
        cout<< absSumDidd(arr,n);     
    }
    // Contributed by mits

Java

// Java program to find minimum sum of absolute
// differences with an array element.
import java.lang.*;
import java.util.*;
  
public class GFG {
  
    // function to find min sum after operation
    static int absSumDidd(int a[])
    {
        // Sort the array
        Arrays.sort(a);
          
        // Pick middle value
        int midValue = a[a.length / 2];
  
        // Sum of absolute differences.
        int sum = 0;
        for (int i = 0; i < a.length; i++) {
            sum = sum + Math.abs(a[i] - midValue);
        }
         
        return sum;       
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 5, 11, 14, 10, 17, 15 };
        System.out.print(absSumDidd(arr));        
    }
    // Contributed by Saurav Jain
}

Python3

# Python3 program to find minimum sum of 
# absolute differences with an array element. 
  
# function to find min sum after operation 
def absSumDidd(a, n): 
      
    # Sort the array 
    a.sort() 
      
    # Pick middle value 
    midValue = a[(int)(n // 2)] 
  
    # Sum of absolute differences. 
    sum = 0
    for i in range(n): 
        sum = sum + abs(a[i] - midValue)
    return sum
  
# Driver Code 
arr = [5, 11, 14, 10, 17, 15] 
n = len(arr)
print(absSumDidd(arr, n))
  
# This code is contributed 
# by sahishelangia

C#

// C# program to find minimum sum of absolute 
// differences with an array element. 
using System;
  
  
 class GFG { 
    
    // function to find min sum after operation 
    static int absSumDidd(int []a) 
    { 
        // Sort the array 
        Array.Sort(a); 
            
        // Pick middle value 
        int midValue = a[a.Length / 2]; 
    
        // Sum of absolute differences. 
        int sum = 0; 
        for (int i = 0; i < a.Length; i++) { 
            sum = sum + Math.Abs(a[i] - midValue); 
        } 
           
        return sum;        
    } 
    
    // Driver Code 
    public static void Main() 
    { 
        int []arr = { 5, 11, 14, 10, 17, 15 }; 
        Console.Write(absSumDidd(arr));         
    } 
    // Contributed by Subhadeep
}

输出：

20

时间复杂度： O(n Log n)

通过使用线性时间算法进行中值查找，我们可以进一步优化上述对O(n)的解。