Dada una array arr[] y un entero K , la tarea es eliminar K % por ciento de los elementos de array de los elementos de array más pequeños y más grandes y calcular la media de la array restante.
Ejemplos:
Entrada: arr[] = {6, 2, 7, 5, 1, 2, 0, 3, 10, 2, 5, 0, 5, 5, 0, 8, 7, 6, 8, 0}, K = 5
Salida: 4.00000
Explicación:
Hay 20 elementos en la array. Por lo tanto, el 5% de 20 es 1 . Por lo tanto, se elimina 1 de los elementos más pequeños (es decir, 0) y 1 elemento de los elementos más grandes (es decir, 0) . Por lo tanto, la media de la array restante es 18 .
Entrada: arr[] = {6, 0, 7, 0, 7, 5, 7, 8, 3, 4, 0, 7, 8, 1, 6, 8, 1, 1, 2, 4}, K = 10
Salida: 4.31250
Acercarse:
- Ordene la array arr[] .
- Encuentre el tamaño de la array .
- Calcule el K -ésimo por ciento del tamaño de la array.
- Ahora, agregue los elementos presentes en los índices K% a (N – 1) – K% .
- Finalmente, encuentre la media de esos elementos.
A continuación se muestra la implementación del enfoque anterior:
C++14
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate the mean
// of a given array after removal
// of Kth percent of smallest and
// largest array elements
void meanOfRemainingElements(int arr[],
int N, int K)
{
// Sort the array
sort(arr, arr + N);
// Find the K-th percent
// of the array size
int kthPercent = (N * K) / 100;
float sum = 0;
// Traverse the array
for (int i = 0; i < N; i++)
// Skip the first K-th
// percent & last K-th
// percent array elements
if (i >= kthPercent && i < (N - kthPercent))
sum += arr[i];
// Mean of the rest of elements
float mean = sum
/ (N - 2 * kthPercent);
// Print mean upto 5 decimal places
cout << fixed << setprecision(5) << mean << endl;
}
// Driver Code
int main()
{
int arr[] = { 6, 2, 7, 5, 1, 2, 0, 3, 10, 2,
5, 0, 5, 5, 0, 8, 7, 6, 8, 0 };
int arr_size = sizeof(arr) / sizeof(arr[0]);
int K = 5;
meanOfRemainingElements(arr, arr_size, K);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
public class GFG
{
// Function to calculate the mean
// of a given array after removal
// of Kth percent of smallest and
// largest array elements
static void meanOfRemainingElements(int[] arr, int N,
int K)
{
// Sort the array
Arrays.sort(arr);
// Find the K-th percent
// of the array size
int kthPercent = (N * K) / 100;
float sum = 0f;
// Traverse the array
for (int i = 0; i < N; i++)
// Skip the first K-th
// percent & last K-th
// percent array elements
if (i >= kthPercent && i < (N - kthPercent))
sum += arr[i];
// Mean of the rest of elements
float mean = (sum / (N - 2 * kthPercent));
// Print mean upto 5 decimal places
System.out.format("%.5f", mean);
}
// Driver Code
public static void main(String args[])
{
int[] arr = { 6, 2, 7, 5, 1, 2, 0, 3, 10, 2,
5, 0, 5, 5, 0, 8, 7, 6, 8, 0 };
int arr_size = arr.length;
int K = 5;
meanOfRemainingElements(arr, arr_size, K);
}
}
// This code is contributed by jana_sayantan.
Python3
# Python program for the above approach # Function to calculate the mean # of a given array after removal # of Kth percent of smallest and # largest array elements def meanOfRemainingElements(arr, N, K): # Sort the array arr.sort() # Find the K-th percent # of the array size kthPercent = (N * K) / 100 sum = 0 # Traverse the array for i in range(N): # Skip the first K-th # percent & last K-th # percent array elements if (i >= kthPercent and i < (N - kthPercent)): sum += arr[i] # Mean of the rest of elements mean = sum/ (N - 2 * kthPercent) # Print mean upto 5 decimal places print( '%.5f'%mean) # Driver Code arr = [ 6, 2, 7, 5, 1, 2, 0, 3, 10, 2, 5, 0, 5, 5, 0, 8, 7, 6, 8, 0 ] arr_size = len(arr) K = 5 meanOfRemainingElements(arr, arr_size, K) # This code is contributed by rohitsingh07052.
C#
// C# program for the above approach
using System;
class GFG {
// Function to calculate the mean
// of a given array after removal
// of Kth percent of smallest and
// largest array elements
static void meanOfRemainingElements(int[] arr, int N,
int K)
{
// Sort the array
Array.Sort(arr);
// Find the K-th percent
// of the array size
int kthPercent = (N * K) / 100;
float sum = 0f;
// Traverse the array
for (int i = 0; i < N; i++)
// Skip the first K-th
// percent & last K-th
// percent array elements
if (i >= kthPercent && i < (N - kthPercent))
sum += arr[i];
// Mean of the rest of elements
float mean = (sum / (N - 2 * kthPercent));
// Print mean upto 5 decimal places
Console.WriteLine(Math.Round(mean,5));
}
// Driver Code
public static void Main()
{
int[] arr = { 6, 2, 7, 5, 1, 2, 0, 3, 10, 2,
5, 0, 5, 5, 0, 8, 7, 6, 8, 0 };
int arr_size = arr.Length;
int K = 5;
meanOfRemainingElements(arr, arr_size, K);
}
}
// This code is contributed by chitranayal.
Javascript
<script>
// Javascript program for the above approach
// Function to calculate the mean
// of a given array after removal
// of Kth percent of smallest and
// largest array elements
function meanOfRemainingElements( arr, N, K)
{
// Sort the array
arr.sort(function(a, b){return a-b});
// Find the K-th percent
// of the array size
let kthPercent = Math.floor((N * K) / 100);
let sum = 0;
// Traverse the array
for (let i = 0; i < N; i++)
// Skip the first K-th
// percent & last K-th
// percent array elements
if (i >= kthPercent && i < (N - kthPercent))
sum += arr[i];
// Mean of the rest of elements
let mean = sum
/ (N - 2 * kthPercent);
// Print mean upto 5 decimal places
document.write(mean.toFixed(5));
}
// Driver Code
let arr = [ 6, 2, 7, 5, 1, 2, 0, 3, 10, 2,
5, 0, 5, 5, 0, 8, 7, 6, 8, 0 ];
let arr_size = arr.length;
let K = 5;
meanOfRemainingElements(arr, arr_size, K);
</script>
Producción:
4.00000
Complejidad de tiempo: O(N * logN)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por sayantanbose2001 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA