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# Average of remaining elements after removing K largest and K smallest elements from array

• Difficulty Level : Basic
• Last Updated : 08 Apr, 2021

Given an array of N integers. The task is to find the average of the numbers after removing k largest elements and k smallest element from the array i.e. calculate the average value of the remaining N – 2K elements.

Examples:

Input: arr = [1, 2, 4, 4, 5, 6], K = 2
Output: 4
Remove 2 smallest elements i.e. 1 and 2
Remove 2 largest elements i.e. 5 and 6
Remaining elements are 4, 4. So average of 4, 4 is 4.

Input: arr = [1, 2, 3], K = 3
Output: 0

Approach:

• If no. of elements to be removed is greater than no. of elements present in the array, then ans = 0.
• Else, Sort all the elements of the array. Then, calculate average of elements from Kth index to n-k-1th index.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the above approach#include using namespace std; // Function to find averagedouble average(int arr[], int n, int k){    double total = 0;     // base case if 2*k>=n    // means all element get removed    if (2 * k >= n)        return 0;     // first sort all elements    sort(arr, arr + n);    int start = k, end = n - k - 1;     // sum of req number    for (int i = start; i <= end; i++)        total += arr[i];     // find average    return (total / (n - 2 * k));} // Driver codeint main(){    int arr[] = { 1, 2, 4, 4, 5, 6 };    int n = sizeof(arr) / sizeof(arr[0]);    int k = 2;     cout << average(arr, n, k) << endl;     return 0;}

## Java

 // Java implementation of the above approach import java.io.*;import java.util.*;class GFG { // Function to find averagestatic double average(int arr[], int n, int k){    double total = 0;     // base case if 2*k>=n    // means all element get removed    if (2 * k >= n)        return 0;     // first sort all elements    Arrays.sort(arr);    int start = k, end = n - k - 1;     // sum of req number    for (int i = start; i <= end; i++)        total += arr[i];     // find average    return (total / (n - 2 * k));} // Driver code      public static void main (String[] args) {            int arr[] = { 1, 2, 4, 4, 5, 6 };    int n = arr.length;    int k = 2;     System.out.println( average(arr, n, k));     }}// This code is contributed by anuj_67..

## Python3

 # Python3 implementation of the# above approach # Function to find averagedef average(arr, n, k) :    total = 0     # base case if 2*k>=n    # means all element get removed    if (2 * k >= n) :        return 0     # first sort all elements    arr.sort()         start , end = k , n - k - 1     # sum of req number    for i in range(start, end + 1) :        total += arr[i]     # find average    return (total / (n - 2 * k)) # Driver codeif __name__ == "__main__" :     arr = [ 1, 2, 4, 4, 5, 6 ]    n = len(arr)    k = 2     print(average(arr, n, k)) # This code is contributed by Ryuga

## C#

 // C# implementation of the above approach using System;public class GFG {      // Function to find average    static double average(int []arr, int n, int k)    {        double total = 0;         // base case if 2*k>=n        // means all element get removed        if (2 * k >= n)            return 0;         // first sort all elements        Array.Sort(arr);        int start = k, end = n - k - 1;         // sum of req number        for (int i = start; i <= end; i++)            total += arr[i];         // find average        return (total / (n - 2 * k));    }     // Driver code          public static void Main() {                int []arr = { 1, 2, 4, 4, 5, 6 };        int n = arr.Length;        int k = 2;         Console.WriteLine( average(arr, n, k));     }}//This code is contributed by 29AjayKumar

## PHP

 =n    // means all element get removed    if (2 * \$k >= \$n)        return 0;     // first sort all elements    sort(\$arr) ;         \$start = \$k ;    \$end = \$n - \$k - 1;     // sum of req number    for (\$i = \$start; \$i <= \$end; \$i++)        \$total += \$arr[\$i];     // find average    return (\$total / (\$n - 2 * \$k));} // Driver code\$arr = array(1, 2, 4, 4, 5, 6);\$n = sizeof(\$arr);\$k = 2; echo average(\$arr, \$n, \$k); // This code is contributed by Ryuga?>

## Javascript


Output:
4

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