Find K elements whose absolute difference with median of array is maximum

Difficulty Level : Medium
Last Updated : 14 Jul, 2020

Given an array arr[] and an integer K, the task is to find the K elements of the array whose absolute difference with median of array is maximum.
Note: If two elements have equal difference then the maximum element is taken into consideration.
Examples:

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}

Approach:

Sort the array and find the median of the array
Create a difference array to store the difference of each element with the median of the sorted array.
Highest difference elements will be the corner elements of the array. Therefore, intialize the two pointers as both the corner elements of the array that is 0 and N – 1.
Finally include the elements of the array one by one with the maximum difference with the median.

Below is the implementation of the above approach:

C++

// C++ implementation to find first K 
// elements whose difference with the  
// median of array is maximum 
  
#include <bits/stdc++.h> 
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 

Output:

5 1 4

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