具有最小绝对差之和的数组元素

给定N个整数的数组arr [] ，任务是从数组中查找元素x ，使得| arr [0] – x | + | arr [1] – x | + | arr [2] – x | +…+ | arr [n – 1] – x |最小化，然后打印最小化的总和。
例子：

Input: arr[] = {1, 3, 9, 3, 6} 
Output: 11 
The optimal solution is to choose x = 3, which produces the sum 
|1 – 3| + |3 – 3| + |9 – 3| + |3 – 3| + |6 – 3| = 2 + 0 + 6 + 0 + 3 = 11
Input: arr[] = {1, 2, 3, 4} 
Output: 4 
 

一个简单的解决方案是遍历每个元素，并检查是否给出了最佳解决方案。该解决方案的时间复杂度为O(n * n)
一种有效的方法：总是选择x作为数组的中位数。如果n是偶数，并且有两个中位数，则两个中位数都是最佳选择。该方法的时间复杂度为O(n * log(n))，因为必须对数组进行排序才能找到中值。当找到x(数组的中位数)时，计算并打印最小化的总和。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the minimized sum
int minSum(int arr[], int n)
{
    // Sort the array
    sort(arr, arr + n);
 
    // Median of the array
    int x = arr[n / 2];
 
    int sum = 0;
 
    // Calculate the minimized sum
    for (int i = 0; i < n; i++)
        sum += abs(arr[i] - x);
 
    // Return the required sum
    return sum;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 3, 9, 3, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << minSum(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function to return the minimized sum
static int minSum(int arr[], int n)
{
    // Sort the array
    Arrays.sort(arr);
 
    // Median of the array
    int x = arr[(int)n / 2];
 
    int sum = 0;
 
    // Calculate the minimized sum
    for (int i = 0; i < n; i++)
        sum += Math.abs(arr[i] - x);
 
    // Return the required sum
    return sum;
}
 
// Driver code
public static void main(String args[])
{
    int arr[] = { 1, 3, 9, 3, 6 };
    int n = arr.length;
    System.out.println(minSum(arr, n));
}
}
 
// This code is contribute by
// Surendra_Gangwar

Python3

# Python3 implementation of the approach
 
# Function to return the minimized sum
def minSum(arr, n) :
     
    # Sort the array
    arr.sort();
 
    # Median of the array
    x = arr[n // 2];
 
    sum = 0;
 
    # Calculate the minimized sum
    for i in range(n) :
        sum += abs(arr[i] - x);
 
    # Return the required sum
    return sum;
 
# Driver code
if __name__ == "__main__" :
     
    arr = [ 1, 3, 9, 3, 6 ];
    n = len(arr)
    print(minSum(arr, n));
 
# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the minimized sum
static int minSum(int []arr, int n)
{
    // Sort the array
    Array.Sort(arr);
 
    // Median of the array
    int x = arr[(int)(n / 2)];
 
    int sum = 0;
 
    // Calculate the minimized sum
    for (int i = 0; i < n; i++)
        sum += Math.Abs(arr[i] - x);
 
    // Return the required sum
    return sum;
}
 
// Driver code
static void Main()
{
    int []arr = { 1, 3, 9, 3, 6 };
    int n = arr.Length;
    Console.WriteLine(minSum(arr, n));
}
}
 
// This code is contributed by mits

Javascript

 
//Javascript implementation of the approach
 
 
// Function to return the minimized sum
function minSum(arr, n)
{
    // Sort the array
    arr.sort();
 
    // Median of the array
    let x = arr[Math.floor(n / 2)];
 
    let sum = 0;
 
    // Calculate the minimized sum
    for (let i = 0; i < n; i++)
        sum += Math.abs(arr[i] - x);
 
    // Return the required sum
    return sum;
}
 
// Driver code
    let arr = [ 1, 3, 9, 3, 6 ];
    let n = arr.length;
    document.write(minSum(arr, n));
 
// This code is contributed by Mayank Tyagi
 

输出：

11

上述解决方案的时间复杂度为O(n Log n)。我们可以使用线性时间算法进一步优化它以在O(n)中工作，以找到第k个最大元素。

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