从与所有其他点的距离总和最小的给定 N 个点中找到 X 轴上的点

给定一个由N 个整数组成的数组arr[] ，表示位于X 轴上的N个点，任务是找到与所有其他点的距离之和最小的点。

例子：

Input: arr[] = {4, 1, 5, 10, 2} 
Output: (4, 0) 
Explanation: 
Distance of 4 from rest of the elements = |4 – 1| + |4 – 5| + |4 – 10| + |4 – 2| = 12 
Distance of 1 from rest of the elements = |1 – 4| + |1 – 5| + |1 – 10| + |1 – 2| = 17 
Distance of 5 from rest of the elements = |5 – 1| + |5 – 4| + |5 – 2| + |5 – 10| = 13 
Distance of 10 from rest of the elements = |10 – 1| + |10 – 2| + |10 – 5| + |10 – 4| = 28 
Distance of 2 from rest of the elements = |2 – 1| + |2 – 4| + |2 – 5| + |2 – 10| = 14
Input: arr[] = {3, 5, 7, 10} 
Output: 5 
 

天真的方法：
任务是遍历数组，并为每个数组元素计算其与所有其他数组元素的绝对差之和。最后，打印具有最大差异总和的数组元素。
时间复杂度： O(N ² )
辅助空间： O(1)

高效的方法：为了优化上述方法，其思想是找到数组的中位数。数组的中位数与数组中其他元素的总距离尽可能小。对于具有偶数个元素的数组，有两个可能的中位数，并且两者的总距离相同，返回具有较低索引的那个，因为它更接近原点。

请按照以下步骤解决问题：

• 对给定的数组进行排序。
• 如果N是奇数，则返回第(N + 1 / 2)^个元素。
• 否则，返回第(N / 2)^个元素。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to find median of the array
int findLeastDist(int A[], int N)
{
    // Sort the given array
    sort(A, A + N);
 
    // If number of elements are even
    if (N % 2 == 0) {
 
        // Return the first median
        return A[(N - 1) / 2];
    }
 
    // Otherwise
    else {
        return A[N / 2];
    }
}
 
// Driver Code
int main()
{
 
    int A[] = { 4, 1, 5, 10, 2 };
    int N = sizeof(A) / sizeof(A[0]);
    cout << "(" << findLeastDist(A, N)
         << ", " << 0 << ")";
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to find median of the array
static int findLeastDist(int A[], int N)
{
     
    // Sort the given array
    Arrays.sort(A);
 
    // If number of elements are even
    if (N % 2 == 0)
    {
         
        // Return the first median
        return A[(N - 1) / 2];
    }
 
    // Otherwise
    else
    {
        return A[N / 2];
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int A[] = { 4, 1, 5, 10, 2 };
    int N = A.length;
     
    System.out.print("(" + findLeastDist(A, N) +
                    ", " + 0 + ")");
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to implement
# the above approach
 
# Function to find median of the array
def findLeastDist(A, N):
     
    # Sort the given array
    A.sort();
 
    # If number of elements are even
    if (N % 2 == 0):
 
        # Return the first median
        return A[(N - 1) // 2];
 
    # Otherwise
    else:
        return A[N // 2];
 
# Driver Code
A = [4, 1, 5, 10, 2];
N = len(A);
 
print("(" , findLeastDist(A, N),
      ", " , 0 , ")");
 
# This code is contributed by PrinciRaj1992

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to find median of the array
static int findLeastDist(int []A, int N)
{
     
    // Sort the given array
    Array.Sort(A);
 
    // If number of elements are even
    if (N % 2 == 0)
    {
         
        // Return the first median
        return A[(N - 1) / 2];
    }
 
    // Otherwise
    else
    {
        return A[N / 2];
    }
}
 
// Driver Code
public static void Main(string[] args)
{
    int []A = { 4, 1, 5, 10, 2 };
    int N = A.Length;
     
    Console.Write("(" + findLeastDist(A, N) +
                 ", " + 0 + ")");
}
}
 
// This code is contributed by rutvik_56

Javascript

 
// Javascript Program to implement
// the above approach
 
// Function to find median of the array
function findLeastDist(A, N)
{
    // Sort the given array
    A.sort((a,b) => a-b);
 
    console.log(A);
     
    // If number of elements are even
    if ((N % 2) == 0) {
 
        // Return the first median
        return A[parseInt((N - 1) / 2)];
    }
 
    // Otherwise
    else {
        return A[parseInt(N / 2)];
    }
}
 
// Driver Code
var A = [ 4, 1, 5, 10, 2 ];
var N = A.length;
document.write( "(" + findLeastDist(A, N)
     + ", " + 0 + ")");
 

输出：

(4, 0)

时间复杂度： O(Nlog(N))
辅助空间： O(1)