Ordene una array de acuerdo con el recuento creciente de factores primos distintos

Dada una array de enteros. La tarea es ordenar la array dada en función del número creciente de factores primos distintos.
Ejemplos: 

Input : arr[] = {30, 2, 1024, 210, 3, 6}
Output : 2 1024 3 6 30 210 

Input : arr[] = {12, 16, 27, 6}
Output : 16 27 6 12

Un enfoque ingenuo es encontrar todos los factores primos de cada elemento de la array y emparejar la cuenta de los factores primos con el elemento en un vector y clasificar la array con respecto a la cuenta de factores primos. 

Un enfoque eficiente es usar un tamiz para encontrar el conteo de distintos factores primos y almacenarlos en un vector. Ahora, recorra la array y empareje el conteo de factores primos distintos con el elemento en un vector y ordene la array con respecto al conteo de factores primos usando una función de comparación .

A continuación se muestra la implementación de este enfoque:  

C++

// C++ program to sort array according to the
// count of distinct prime factors
#include <bits/stdc++.h>
using namespace std;
 
// array to store the count of distinct prime
int prime[100001];
 
void SieveOfEratosthenes()
{
    // Create a int array "prime[0..n]" and initialize
    // all entries it as 0. A value in prime[i] will
    // count of distinct prime factors.
 
    memset(prime, 0, sizeof(prime));
 
    // 0 and 1 does not have any prime factors
    prime[0] = 0;
    prime[1] = 0;
 
    for (int p = 2; p * p <= 100001; p++) {
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == 0) {
            prime[p]++;
 
            // Update all multiples of p greater than or
            // equal to the square of it
            // numbers which are multiple of p and are
            // less than p^2 are already been marked.
            for (int i = p * p; i <= 100001; i += p)
                prime[i]++;
        }
    }
}
 
// comparator function to sort the vector in
// ascending order of second element of the pair
bool Compare(pair<int, int> p1, pair<int, int> p2)
{
    return (p1.second < p2.second);
}
 
// Function to sort the array on the
// basis increasing count of distinct
// prime factors
void sortArr(int arr[], int n)
{
    // vector to store the number and
    // count of prime factor
    vector<pair<int, int> > v;
 
    for (int i = 0; i < n; i++) {
        // push_back the element and
        // count of distinct
        // prime factors
        v.push_back(make_pair(arr[i], prime[arr[i]]));
    }
 
    // sort the array on the
    // basis increasing count of
    // distinct prime factors
    sort(v.begin(), v.end(), Compare);
 
    // display the sorted array
    for (int i = 0; i < n; i++)
        cout << v[i].first << " ";
 
    cout << endl;
}
 
// Driver code
int main()
{
    // create the sieve
    SieveOfEratosthenes();
 
    int arr[] = { 30, 2, 1024, 210, 3, 6 };
 
    int n = sizeof(arr) / sizeof(int);
 
    sortArr(arr, n);
 
    return 0;
}

Java

import java.util.Arrays;
 
class GFG
{
     
    static class Pair implements Comparable<Pair>
    {
        int first, second;
         
        Pair(int f, int s)
        {
            first = f;
            second = s;
        }
 
        @Override
        public int compareTo(Pair o)
        {
            if(this.second > o.second)
                return 1;
            else if(this.second == o.second)
                return 0;
            return -1;
        }
    }
    // array to store the count of distinct prime
    static int prime[] = new int[100002];
     
    static void SieveOfEratosthenes()
    {
        // Create a int array "prime[0..n]" and initialize
        // all entries it as 0. A value in prime[i] will
        // count of distinct prime factors.
        Arrays.fill(prime, 0);
     
        // 0 and 1 does not have any prime factors
        prime[0] = 0;
        prime[1] = 0;
     
        for (int p = 2; p * p <= 100001; p++)
        {
            // If prime[p] is not changed,
            // then it is a prime
            if (prime[p] == 0)
            {
                prime[p]++;
     
                // Update all multiples of p greater than or
                // equal to the square of it
                // numbers which are multiple of p and are
                // less than p^2 are already been marked.
                for (int i = p * p; i <= 100001; i += p)
                    prime[i]++;
            }
        }
    }
     
    // Function to sort the array on the
    // basis increasing count of distinct
    // prime factors
    static void sortArr(int arr[], int n)
    {
        // Array to store the number and
        // count of prime factor
        Pair v[] = new Pair[n];
     
        for (int i = 0; i < n; i++)
        {
            // update the element and
            // count of distinct
            // prime factors
            v[i] = new Pair(arr[i], prime[arr[i]]);
        }
     
        // sort the array on the
        // basis increasing count of
        // distinct prime factors
        Arrays.sort(v);
     
        // display the sorted array
        for (int i = 0; i < n; i++)
            System.out.print(v[i].first + " ");
     
        System.out.println();
    }
     
    // Driver code
    public static void main(String args[])
    {
        // create the sieve
        SieveOfEratosthenes();
     
        int arr[] = { 30, 2, 1024, 210, 3, 6 };
     
        int n = arr.length;
     
        sortArr(arr, n);
    }
}
 
// This code is contributed by ghanshyampandey

Python3

# Python3 program to sort array according to the
# count of distinct prime factors
import functools as ft
 
# array to store the count of distinct prime
prime = [0 for i in range(100001)]
 
def SieveOfEratosthenes():
 
    # Create a array "prime[0..n]" and initialize
    # all entries it as 0. A value in prime[i]
    # will count of distinct prime factors.
 
    # memset(prime, 0, sizeof(prime))
 
    # 0 and 1 does not have any prime factors
    prime[0] = 0
    prime[1] = 0
 
    for p in range(2, 100002):
 
        if p * p > 100001:
            break
 
        # If prime[p] is not changed,
        # then it is a prime
        if (prime[p] == 0):
            prime[p] += 1
 
            # Update all multiples of p greater than
            # or equal to the square of it
            # numbers which are multiple of p and are
            # less than p^2 are already been marked.
            for i in range(2 * p, 100001, p):
                prime[i] += 1
 
# Function to sort the array on the
# basis increasing count of distinct
# prime factors
def sortArr(arr, n):
 
    # vector to store the number and
    # count of prime factor
    v = []
 
    for i in range(n):
 
        # append the element and
        # count of distinct
        # prime factors
        v.append([arr[i], prime[arr[i]]])
 
    # sort the array on the
    # basis increasing count of
    # distinct prime factors
    v.sort(key= lambda x:x[1])
 
    # display the sorted array
 
    for i in range(n):
        print(v[i][0], end = " ")
 
    print()
 
# Driver code
 
# create the sieve
SieveOfEratosthenes()
 
arr = [30, 2, 1024, 210, 3, 6]
 
n = len(arr)
 
sortArr(arr, n)
 
# This code is contributed by Mohit Kumar

C#

// C# program to sort array according to the
// count of distinct prime factors
using System;
using System.Collections.Generic;
 
class GFG{
 
public class Pair : IComparable<Pair>
{
    public int first, second;
 
    public Pair(int f, int s)
    {
        first = f;
        second = s;
    }
 
    // @Override
    public int CompareTo(Pair o)
    {
        if (this.second > o.second)
            return 1;
        else if (this.second == o.second)
            return 0;
             
        return -1;
    }
}
 
// Array to store the count of distinct prime
static int[] prime = new int[100002];
 
static void SieveOfEratosthenes()
{
     
    // Create a int array "prime[0..n]"
    // and initialize all entries it as
    // 0. A value in prime[i] will
    // count of distinct prime factors.
    Array.Fill(prime, 0);
 
    // 0 and 1 does not have any prime factors
    prime[0] = 0;
    prime[1] = 0;
 
    for(int p = 2; p * p <= 100001; p++)
    {
         
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == 0)
        {
            prime[p]++;
 
            // Update all multiples of p greater
            // than or equal to the square of it
            // numbers which are multiple of p and are
            // less than p^2 are already been marked.
            for(int i = p * p; i <= 100001; i += p)
                prime[i]++;
        }
    }
}
 
// Function to sort the array on the
// basis increasing count of distinct
// prime factors
static void sortArr(int[] arr, int n)
{
     
    // Array to store the number and
    // count of prime factor
    Pair[] v = new Pair[n];
 
    for(int i = 0; i < n; i++)
    {
         
        // Update the element and
        // count of distinct
        // prime factors
        v[i] = new Pair(arr[i],
                  prime[arr[i]]);
    }
 
    // Sort the array on the
    // basis increasing count of
    // distinct prime factors
    Array.Sort(v);
 
    // Display the sorted array
    for(int i = 0; i < n; i++)
        Console.Write(v[i].first + " ");
 
    Console.WriteLine();
}
 
// Driver code
public static void Main(string[] args)
{
     
    // Create the sieve
    SieveOfEratosthenes();
 
    int[] arr = { 30, 2, 1024, 210, 3, 6 };
    int n = arr.Length;
 
    sortArr(arr, n);
}
}
 
// This code is contributed by grand_master

Javascript

<script>
 
   // JavaScript program to sort array according to the
  // count of distinct prime factors
     
    // array to store the count of distinct prime
    let prime = new Array(100002);
     
    function SieveOfEratosthenes()
    {
        // Create a int array "prime[0..n]" and initialize
        // all entries it as 0. A value in prime[i] will
        // count of distinct prime factors.
        for(let i=0;i<prime.length;i++)
        {
            prime[i]=0;
        }
      
        // 0 and 1 does not have any prime factors
        prime[0] = 0;
        prime[1] = 0;
      
        for (let p = 2; p * p <= 100001; p++)
        {
            // If prime[p] is not changed,
            // then it is a prime
            if (prime[p] == 0)
            {
                prime[p]++;
      
                // Update all multiples of p greater than or
                // equal to the square of it
                // numbers which are multiple of p and are
                // less than p^2 are already been marked.
                for (let i = p * p; i <= 100001; i += p)
                    prime[i]++;
            }
        }
    }
     
    // Function to sort the array on the
    // basis increasing count of distinct
    // prime factors
    function sortArr(arr,n)
    {
        // Array to store the number and
        // count of prime factor
        let v=[];
      
        for (let i = 0; i < n; i++)
        {
            // update the element and
            // count of distinct
            // prime factors
            v.push([arr[i], prime[arr[i]]]);
             
        }
         
         
      
        // sort the array on the
        // basis increasing count of
        // distinct prime factors
        v.sort(function(a,b){return a[1]-b[1];});
      
        // display the sorted array
        for (let i = 0; i < n; i++)
            document.write(v[i][0] + " ");
      
        document.write("<br>");
    }
     
    // Driver code
     
    // create the sieve
    SieveOfEratosthenes();
     
    let arr=[30, 2, 1024, 210, 3, 6];
    let n = arr.length;
     
    sortArr(arr, n);
 
     
 
// This code is contributed by patel2127
 
</script>
Producción: 

2 1024 3 6 30 210

 

Publicación traducida automáticamente

Artículo escrito por andrew1234 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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