Ordenar números feos en una array en sus posiciones relativas

Dada una array de enteros arr[] , la tarea es ordenar solo aquellos elementos que son números feos en sus posiciones relativas en la array (las posiciones de otros elementos no deben verse afectadas). 
Los números feos son números cuyos únicos factores primos son 2 , 3 o 5 . 
La secuencia 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ….. muestra los primeros números feos. Por convención, 1 está incluido.
Ejemplos: 
 

Entrada: arr[] = {13, 9, 11, 3, 2} 
Salida: 13 2 11 3 9 
9, 3 y 2 son los únicos números feos en la array dada.
Entrada: arr[] = {1, 2, 3, 7, 12, 10} 
Salida: 1 2 3 7 10 12 
 

Acercarse: 
 

• Comience a recorrer la array y para cada elemento arr[i] , si arr[i] es un número feo, guárdelo en una ArrayList y actualice arr[i] = -1
• Después de que todos los números desagradables se hayan almacenado en ArrayList, ordene el ArrayList actualizado.
• Recorra nuevamente la array y para cada elemento, 
  • Si arr[i] = -1 , imprima el primer elemento de ArrayList que no se haya impreso antes.
  • De lo contrario, imprime arr[i] .

A continuación se muestra la implementación del enfoque anterior:
 

// CPP implementation of the approach
#include<bits/stdc++.h>
using namespace std;
 
// Function that returns true if n is an ugly number
bool isUgly(int n)
{
    // While divisible by 2, keep dividing
    while (n % 2 == 0)
        n = n / 2;
 
    // While divisible by 3, keep dividing
    while (n % 3 == 0)
        n = n / 3;
 
    // While divisible by 5, keep dividing
    while (n % 5 == 0)
        n = n / 5;
 
    // n must be 1 if it was ugly
    if (n == 1)
        return true;
    return false;
}
 
// Function to sort ugly numbers
// in their relative positions
void sortUglyNumbers(int arr[], int n)
{
 
    // To store the ugly numbers from the array
    vector<int> list;
 
    int i;
    for (i = 0; i < n; i++)
    {
 
        // If current element is an ugly number
        if (isUgly(arr[i]))
        {
 
            // Add it to the ArrayList
            // and set arr[i] to -1
            list.push_back(arr[i]);
            arr[i] = -1;
        }
    }
 
    // Sort the ugly numbers
    sort(list.begin(),list.end());
 
    int j = 0;
    for (i = 0; i < n; i++)
    {
 
        // Position of an ugly number
        if (arr[i] == -1)
            cout << list[j++] << " ";
        else
            cout << arr[i] << " ";
    }
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 7, 12, 10 };
    int n = sizeof(arr)/sizeof(arr[0]);
    sortUglyNumbers(arr, n);
}
 
// This code is contributed by
// Surendra_Gangwar

// Java implementation of the approach
import java.util.ArrayList;
import java.util.Collections;
 
class GFG {
 
    // Function that returns true if n is an ugly number
    static boolean isUgly(int n)
    {
        // While divisible by 2, keep dividing
        while (n % 2 == 0)
            n = n / 2;
 
        // While divisible by 3, keep dividing
        while (n % 3 == 0)
            n = n / 3;
 
        // While divisible by 5, keep dividing
        while (n % 5 == 0)
            n = n / 5;
 
        // n must be 1 if it was ugly
        if (n == 1)
            return true;
        return false;
    }
 
    // Function to sort ugly numbers
    // in their relative positions
    static void sortUglyNumbers(int arr[], int n)
    {
 
        // To store the ugly numbers from the array
        ArrayList<Integer> list = new ArrayList<>();
 
        int i;
        for (i = 0; i < n; i++) {
 
            // If current element is an ugly number
            if (isUgly(arr[i])) {
 
                // Add it to the ArrayList
                // and set arr[i] to -1
                list.add(arr[i]);
                arr[i] = -1;
            }
        }
 
        // Sort the ugly numbers
        Collections.sort(list);
 
        int j = 0;
        for (i = 0; i < n; i++) {
 
            // Position of an ugly number
            if (arr[i] == -1)
                System.out.print(list.get(j++) + " ");
            else
                System.out.print(arr[i] + " ");
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 3, 7, 12, 10 };
        int n = arr.length;
        sortUglyNumbers(arr, n);
    }
}

# Python3 implementation of the approach
 
# Function that returns true if n is an ugly number
def isUgly(n):
 
    # While divisible by 2, keep dividing
    while n % 2 == 0:
        n = n // 2
 
    # While divisible by 3, keep dividing
    while n % 3 == 0:
        n = n // 3
 
    # While divisible by 5, keep dividing
    while n % 5 == 0:
        n = n // 5
 
    # n must be 1 if it was ugly
    if n == 1:
        return True
    return False
 
# Function to sort ugly numbers
# in their relative positions
def sortUglyNumbers(arr, n):
 
    # To store the ugly numbers from the array
    list = []
 
    for i in range(0, n):
 
        # If current element is an ugly number
        if isUgly(arr[i]):
 
            # Add it to the ArrayList
            # and set arr[i] to -1
            list.append(arr[i])
            arr[i] = -1
 
    # Sort the ugly numbers
    list.sort()
 
    j = 0
    for i in range(0, n):
     
        # Position of an ugly number
        if arr[i] == -1:
            print(list[j], end = " ")
            j += 1
        else:
            print(arr[i], end = " ")
 
# Driver code
if __name__ == "__main__":
 
    arr = [1, 2, 3, 7, 12, 10]
    n = len(arr)
    sortUglyNumbers(arr, n)
 
# This code is contributed by Rituraj Jain

// C# implementation of the approach
using System;
using System.Collections.Generic;
     
class GFG
{
 
    // Function that returns true
    // if n is an ugly number
    static bool isUgly(int n)
    {
        // While divisible by 2, keep dividing
        while (n % 2 == 0)
            n = n / 2;
 
        // While divisible by 3, keep dividing
        while (n % 3 == 0)
            n = n / 3;
 
        // While divisible by 5, keep dividing
        while (n % 5 == 0)
            n = n / 5;
 
        // n must be 1 if it was ugly
        if (n == 1)
            return true;
        return false;
    }
 
    // Function to sort ugly numbers
    // in their relative positions
    static void sortUglyNumbers(int []arr, int n)
    {
 
        // To store the ugly numbers from the array
        List<int> list = new List<int>();
 
        int i;
        for (i = 0; i < n; i++)
        {
 
            // If current element is an ugly number
            if (isUgly(arr[i]))
            {
 
                // Add it to the ArrayList
                // and set arr[i] to -1
                list.Add(arr[i]);
                arr[i] = -1;
            }
        }
 
        // Sort the ugly numbers
        list.Sort();
 
        int j = 0;
        for (i = 0; i < n; i++)
        {
 
            // Position of an ugly number
            if (arr[i] == -1)
                Console.Write(list[j++] + " ");
            else
                Console.Write(arr[i] + " ");
        }
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = { 1, 2, 3, 7, 12, 10 };
        int n = arr.Length;
        sortUglyNumbers(arr, n);
    }
}
 
// This code contributed by Rajput-Ji

<script>
 
// Javascript implementation of the approach
 
// Function that returns true if n is an ugly number
function isUgly(n)
{
    // While divisible by 2, keep dividing
    while (n % 2 == 0)
        n = n / 2;
 
    // While divisible by 3, keep dividing
    while (n % 3 == 0)
        n = n / 3;
 
    // While divisible by 5, keep dividing
    while (n % 5 == 0)
        n = n / 5;
 
    // n must be 1 if it was ugly
    if (n == 1)
        return true;
    return false;
}
 
// Function to sort ugly numbers
// in their relative positions
function sortUglyNumbers(arr, n)
{
 
    // To store the ugly numbers from the array
    var list = [];
 
    var i;
    for (i = 0; i < n; i++)
    {
 
        // If current element is an ugly number
        if (isUgly(arr[i]))
        {
 
            // Add it to the ArrayList
            // and set arr[i] to -1
            list.push(arr[i]);
            arr[i] = -1;
        }
    }
 
    // Sort the ugly numbers
    list.sort((a,b)=>a-b);
 
    var j = 0;
    for (i = 0; i < n; i++)
    {
 
        // Position of an ugly number
        if (arr[i] == -1)
            document.write( list[j++] + " ");
        else
            document.write( arr[i] + " ");
    }
}
 
// Driver code
var arr = [1, 2, 3, 7, 12, 10 ];
var n = arr.length;
sortUglyNumbers(arr, n);
 
 
</script>

1 2 3 7 10 12

Complejidad de Tiempo: O(nlog(n))
Espacio Auxiliar: O(n)