Dado un arreglo ‘arr’ de enteros positivos, la tarea es contar el número de números compuestos en el arreglo.
Nota: 1 no es ni primo ni compuesto .
Ejemplos:
Entrada: arr[] = {1, 3, 4, 5, 7}
Salida: 1
4 es el único número compuesto.
Entrada: arr[] = {1, 2, 3, 4, 5, 6, 7}
Salida: 2
Enfoque ingenuo: una solución simple es atravesar la array y hacer una prueba de primalidad en cada elemento.
Enfoque eficiente: el uso de la criba de Eratóstenes genera un vector booleano hasta el tamaño del elemento máximo de la array que se puede usar para verificar si un número es primo o no. También agregue 0 y 1 como números primos para que no se cuenten como números compuestos. Ahora recorra la array y encuentre el recuento de los elementos que están compuestos utilizando el vector booleano generado.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to count the
// number of composite numbers
// in the given array
#include <bits/stdc++.h>
using namespace std;
// Function that returns the
// the count of composite numbers
int compositeCount(int arr[], int n, int* sum)
{
// Find maximum value in the array
int max_val = *max_element(arr, arr + n);
// Use sieve to find all prime numbers
// less than or equal to max_val
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
vector<bool> prime(max_val + 1, true);
// Set 0 and 1 as primes as
// they don't need to be
// counted as composite numbers
prime[0] = true;
prime[1] = true;
for (int p = 2; p * p <= max_val; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= max_val; i += p)
prime[i] = false;
}
}
// Count all composite
// numbers in the arr[]
int count = 0;
for (int i = 0; i < n; i++)
if (!prime[arr[i]]) {
count++;
*sum = *sum + arr[i];
}
return count;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
int sum = 0;
cout << "Count of Composite Numbers = "
<< compositeCount(arr, n, &sum);
cout << "\nSum of Composite Numbers = " << sum;
return 0;
}
Java
import java.util.*;
// Java program to count the
// number of composite numbers
// in the given array
class GFG
{
static int sum = 0;
// Function that returns the
// the count of composite numbers
static int compositeCount(int arr[], int n)
{
// Find maximum value in the array
int max_val = Arrays.stream(arr).max().getAsInt();
// Use sieve to find all prime numbers
// less than or equal to max_val
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
Vector<Boolean> prime = new Vector<Boolean>(max_val + 1);
for (int i = 0; i < max_val + 1; i++)
{
prime.add(i, Boolean.TRUE);
}
// Set 0 and 1 as primes as
// they don't need to be
// counted as composite numbers
prime.add(0, Boolean.TRUE);
prime.add(1, Boolean.TRUE);
for (int p = 2; p * p <= max_val; p++)
{
// If prime[p] is not changed, then
// it is a prime
if (prime.get(p) == true)
{
// Update all multiples of p
for (int i = p * 2; i <= max_val; i += p)
{
prime.add(i, Boolean.FALSE);
}
}
}
// Count all composite
// numbers in the arr[]
int count = 0;
for (int i = 0; i < n; i++)
{
if (!prime.get(arr[i]))
{
count++;
sum = sum + arr[i];
}
}
return count;
}
// Driver code
public static void main(String[] args)
{
int arr[] = {1, 2, 3, 4, 5, 6, 7};
int n = arr.length;
System.out.print("Count of Composite Numbers = "
+ compositeCount(arr, n));
System.out.print("\nSum of Composite Numbers = " + sum);
}
}
// This code has been contributed by 29AjayKumar
Python3
# Python3 program to count the
# number of composite numbers
# in the given array
# Function that returns the
# the count of composite numbers
def compositeCount(arr, n):
Sum = 0
# Find maximum value in the array
max_val = max(arr)
# Use sieve to find all prime numbers
# less than or equal to max_val
# Create a boolean array "prime[0..n]".
# A value in prime[i] will finally be
# false if i is Not a prime, else True.
prime = [True for i in range(max_val + 1)]
# Set 0 and 1 as primes as
# they don't need to be
# counted as composite numbers
prime[0] = True
prime[1] = True
for p in range(2, max_val + 1):
if p * p > max_val:
break
# If prime[p] is not changed,
# then it is a prime
if (prime[p] == True):
# Update all multiples of p
for i in range(p * 2, max_val + 1, p):
prime[i] = False
# Count all composite numbers
# in the arr[]
count = 0
for i in range(n):
if (prime[arr[i]] == False):
count += 1
Sum = Sum + arr[i]
return count, Sum
# Driver code
arr = [1, 2, 3, 4, 5, 6, 7 ]
n = len(arr)
count, Sum = compositeCount(arr, n)
print("Count of Composite Numbers = ", count)
print("Sum of Composite Numbers = ", Sum)
// This code is contributed by Mohit Kumar
C#
// C# program to count the
// number of composite numbers
// in the given array
using System;
using System.Linq;
using System.Collections;
class GFG
{
static int sum1=0;
// Function that returns the
// the count of composite numbers
static int compositeCount(int []arr, int n, int sum)
{
// Find maximum value in the array
int max_val = arr.Max();
// Use sieve to find all prime numbers
// less than or equal to max_val
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
bool[] prime=new bool[max_val + 1];
// Set 0 and 1 as primes as
// they don't need to be
// counted as composite numbers
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= max_val; p++)
{
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == false)
{
// Update all multiples of p
for (int i = p * 2; i <= max_val; i += p)
prime[i] = true;
}
}
// Count all composite
// numbers in the arr[]
int count = 0;
for (int i = 0; i < n; i++)
if (prime[arr[i]])
{
count++;
sum = sum + arr[i];
}
sum1 = sum;
return count;
}
// Driver code
static void Main()
{
int []arr = { 1, 2, 3, 4, 5, 6, 7 };
int n = arr.Length;
int sum = 0;
Console.Write("Count of Composite Numbers = "+
compositeCount(arr, n, sum));
Console.Write("\nSum of Composite Numbers = "+sum1);
}
}
// This code is contributed by mits
PHP
<?php
// PHP program to count the
// number of composite numbers
// in the given array
// Function that returns the
// the count of composite numbers
function compositeCount($arr, $n, &$sum)
{
// Find maximum value in the array
$max_val = max($arr);
// Use sieve to find all prime numbers
// less than or equal to max_val
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
$prime=array_fill(0,$max_val + 1, true);
// Set 0 and 1 as primes as
// they don't need to be
// counted as composite numbers
$prime[0] = true;
$prime[1] = true;
for ($p = 2; $p * $p <= $max_val; $p++)
{
// If prime[p] is not changed, then
// it is a prime
if ($prime[$p] == true)
{
// Update all multiples of p
for ($i = $p * 2; $i <= $max_val; $i += $p)
$prime[$i] = false;
}
}
// Count all composite
// numbers in the arr[]
$count = 0;
for ($i = 0; $i < $n; $i++)
if (!$prime[$arr[$i]])
{
$count++;
$sum = $sum + $arr[$i];
}
return $count;
}
// Driver code
$arr = array( 1, 2, 3, 4, 5, 6, 7 );
$n = count($arr);
$sum = 0;
echo "Count of Composite Numbers = ".compositeCount($arr, $n, $sum);
echo "\nSum of Composite Numbers = ".$sum;
// This code is contributed by mits
?>
Javascript
<script>
// Javascript program to count the
// number of composite numbers
// in the given array
// Function that returns the
// the count of composite numbers
let sum = 0;
function compositeCount(arr, n)
{
// Find maximum value in the array
let max_val = arr.sort((a, b) => b - a)[0];
// Use sieve to find all prime numbers
// less than or equal to max_val
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
let prime = new Array(max_val + 1).fill(true);
// Set 0 and 1 as primes as
// they don't need to be
// counted as composite numbers
prime[0] = true;
prime[1] = true;
for (let p = 2; p * p <= max_val; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (let i = p * 2; i <= max_val; i += p)
prime[i] = false;
}
}
// Count all composite
// numbers in the arr[]
let count = 0;
for (let i = 0; i < n; i++)
if (!prime[arr[i]]) {
count++;
sum = sum + arr[i];
}
return count;
}
// Driver code
let arr = new Array(1, 2, 3, 4, 5, 6, 7);
let n = arr.length;
document.write("Count of Composite Numbers = " + compositeCount(arr, n, sum));
document.write("<br>Sum of Composite Numbers = " + sum);
// This code is contributed by gfgking
</script>
Producción:
Count of Composite Numbers = 2 Sum of Composite Numbers = 10
Complejidad del tiempo: O(nlog(logn))
Complejidad del espacio : O(n) ya que se está utilizando el espacio auxiliar
Publicación traducida automáticamente
Artículo escrito por VishalBachchas y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA