查找与数组中位数的绝对差最大的K个元素

给定一个数组arr []和一个整数K ，任务是找到与数组中位数的绝对差最大的数组K个元素。
注意：如果两个元素的差相等，则将最大元素考虑在内。
例子：

Input : arr[] = {1, 2, 3, 4, 5}, k = 3 
Output : {5, 1, 4} 
Explanation : 
Median m = 3, 
Difference of each array elements from median, 
1 ==> diff(1-3) = 2 
2 ==> diff(2-3) = 1 
3 ==> diff(3-3) = 0 
4 ==> diff(4-3) = 1 
5 ==> diff(5-3) = 2 
First K elements are 5, 1, 4 in this array.
Input: arr[] = {1, 2, 3}, K = 2 
Output: {3, 1} 
 

 

方法：

• 对数组进行排序并找到数组的中位数
• 创建一个差值数组以存储每个元素与已排序数组的中位数的差。
• 差异最大的元素将是数组的角元素。因此，将两个指针初始化为数组的两个角元素，即0和N – 1。
• 最后，将数组中的元素与中值最大地一一对应。

下面是上述方法的实现：

C++

// C++ implementation to find first K
// elements whose difference with the 
// median of array is maximum
  
#include 
using namespace std;
  
// Function for calculating median 
double findMedian(int a[], int n) 
{
    // check for even case 
    if (n % 2 != 0) 
       return (double)a[n/2]; 
        
    return (double)(a[(n-1)/2] + a[n/2])/2.0; 
} 
  
// Function to find the K maximum absolute
// difference with the median of the array
void kStrongest(int arr[], int n, int k)
{
    // Sort the array.
    sort(arr, arr + n);
  
    // Store median
    double median = findMedian(arr, n);
    int diff[n];
  
    // Find and store difference
    for (int i = 0; i < n; i++) {
        diff[i] = abs(median - arr[i]);
    }
  
    int i = 0, j = n - 1;
    while (k > 0) {
          
        // If diff[i] is greater print it
        // Else print diff[j]
        if (diff[i] > diff[j]) {
            cout << arr[i] << " ";
            i++;
        }
        else {
            cout << arr[j] << " ";
            j--;
        }
        k--;
    }
}
  
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int k = 3;
    int n = sizeof(arr) / sizeof(arr[0]);
  
    kStrongest(arr, n, k);
    return 0;
}

Java

// Java implementation to find first K
// elements whose difference with the 
// median of array is maximum
import java.util.*;
class GFG{
   
// Function for calculating median 
static double findMedian(int a[], int n) 
{
    // check for even case 
    if (n % 2 != 0) 
       return (double)a[n / 2]; 
         
    return (double)(a[(n - 1) / 2] + 
                    a[n / 2]) / 2.0; 
} 
   
// Function to find the K maximum absolute
// difference with the median of the array
static void kStrongest(int arr[], int n, int k)
{
    // Sort the array.
    Arrays.sort(arr);
   
    // Store median
    double median = findMedian(arr, n);
    int []diff = new int[n];
   
    // Find and store difference
    for (int i = 0; i < n; i++) 
    {
        diff[i] = (int)Math.abs(median - arr[i]);
    }
   
    int i = 0, j = n - 1;
    while (k > 0) 
    {
           
        // If diff[i] is greater print it
        // Else print diff[j]
        if (diff[i] > diff[j])
        {
            System.out.print(arr[i] + " ");
            i++;
        }
        else 
        {
            System.out.print(arr[j] + " ");
            j--;
        }
        k--;
    }
}
   
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int k = 3;
    int n = arr.length;
   
    kStrongest(arr, n, k);
}
}
// This code is contributed by sapnasingh4991

Python3

# Python3 program to find first K
# elements whose difference with the
# median of array is maximum
  
# Function for calculating median
def findMedian(a, n):
      
    # Check for even case
    if (n % 2 != 0):
        return a[int(n / 2)]
          
    return (a[int((n - 1) / 2)] + 
            a[int(n / 2)]) / 2.0
  
# Function to find the K maximum 
# absolute difference with the 
# median of the array
def kStrongest(arr, n, k):
      
    # Sort the array
    arr.sort()
      
    # Store median
    median = findMedian(arr, n)
    diff = [0] * (n)
      
    # Find and store difference
    for i in range(n):
        diff[i] = abs(median - arr[i])
          
    i = 0
    j = n - 1
      
    while (k > 0):
          
        # If diff[i] is greater print 
        # it. Else print diff[j]
        if (diff[i] > diff[j]):
            print(arr[i], end = " ")
            i += 1
        else:
            print(arr[j], end = " ")
            j -= 1
          
        k -= 1
      
# Driver code
arr = [ 1, 2, 3, 4, 5 ]
k = 3
n = len(arr)
  
kStrongest(arr, n, k)
  
# This code is contributed by sanjoy_62

C#

// C# implementation to find first K
// elements whose difference with the 
// median of array is maximum
using System;
  
class GFG{
  
// Function for calculating median 
static double findMedian(int []a, int n) 
{
    // Check for even case 
    if (n % 2 != 0) 
        return (double)a[n / 2]; 
          
    return (double)(a[(n - 1) / 2] + 
                    a[n / 2]) / 2.0; 
} 
  
// Function to find the K maximum absolute
// difference with the median of the array
static void kStrongest(int []arr, int n,
                                  int k)
{
      
    // Sort the array.
    Array.Sort(arr);
      
    int i = 0;
      
    // Store median
    double median = findMedian(arr, n);
    int []diff = new int[n];
  
    // Find and store difference
    for(i = 0; i < n; i++) 
    {
       diff[i] = (int)Math.Abs(median - arr[i]);
    }
  
    int j = n - 1;
    i = 0;
    while (k > 0) 
    {
          
        // If diff[i] is greater print it
        // Else print diff[j]
        if (diff[i] > diff[j])
        {
            Console.Write(arr[i] + " ");
            i++;
        }
        else
        {
            Console.Write(arr[j] + " ");
            j--;
        }
        k--;
    }
}
  
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5 };
    int k = 3;
    int n = arr.Length;
  
    kStrongest(arr, n, k);
}
}
  
// This code is contributed by Rohit_ranjan

输出：

5 1 4