Ordena los números según su producto de dígitos

Dada una array arr[] de N enteros no negativos, la tarea es ordenar estos enteros según el producto de sus dígitos.
Ejemplos: 
 

Entrada: arr[] = {12, 10, 102, 31, 15} 
Salida: 10 102 12 31 15 
10 -> 1 * 0 = 0 
102 -> 1 * 0 * 2 = 0 
12 -> 1 * 2 = 2 
31 -> 3 * 1 = 3 
15 -> 1 * 5 = 5
Entrada: arr[] = {12, 10} 
Salida: 10 12 
 

Enfoque: La idea es almacenar cada elemento con su producto de dígitos en un par de vectores y luego ordenar todos los elementos del vector de acuerdo con los productos de dígitos almacenados. Finalmente, imprima los elementos en orden.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the product
// of the digits of n
int productOfDigit(int n)
{
    int product = 1;
    while (n > 0) {
        product *= n % 10;
        n = n / 10;
    }
    return product;
}
 
// Function to sort the array according to
// the product of the digits of elements
void sortArr(int arr[], int n)
{
    // Vector to store the digit product
    // with respective elements
    vector<pair<int, int> > vp;
 
    // Inserting digit product with elements
    // in the vector pair
    for (int i = 0; i < n; i++) {
        vp.push_back(make_pair(productOfDigit(arr[i]), arr[i]));
    }
 
    // Sort the vector, this will sort the pair
    // according to the product of the digits
    sort(vp.begin(), vp.end());
 
    // Print the sorted vector content
    for (int i = 0; i < vp.size(); i++)
        cout << vp[i].second << " ";
}
 
// Driver code
int main()
{
    int arr[] = { 12, 10, 102, 31, 15 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    sortArr(arr, n);
 
    return 0;
}

# Python3 implementation of the approach
 
# Function to return the product
# of the digits of n
def productOfDigit(n) :
 
    product = 1;
    while (n > 0) :
        product *= (n % 10);
        n = n // 10;
 
    return product;
 
# Function to sort the array according to
# the product of the digits of elements
def sortArr(arr, n) :
     
    # Vector to store the digit product
    # with respective elements
    vp = [];
 
    # Inserting digit product with elements
    # in the vector pair
    for i in range(n) :
        vp.append((productOfDigit(arr[i]), arr[i]));
     
    # Sort the vector, this will sort the pair
    # according to the product of the digits
    vp.sort();
 
    # Print the sorted vector content
    for i in range(len(vp)) :
        print(vp[i][1], end = " ");
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 12, 10, 102, 31, 15 ];
    n = len(arr);
 
    sortArr(arr, n);
 
# This code is contributed by AnkitRai01

<script>
// Javascript implementation of the
// above approach
 
// Function to return the product
// of the digits of n
function productOfDigit(n)
{
    var product = 1;
    while (n > 0) {
        product *= n % 10;
        n = Math.floor(n / 10);
    }
    return product;
}
   
// Function to sort the array according to
// the product of the digits of elements
function sortArr(arr, n)
{
    // Vector to store the digit product
    // with respective elements
    var vp = new Array(n);
    // Loop to create 2D array using 1D array
    for (var i = 0; i < vp.length; i++) {
        vp[i] = [];
    }
   
    // Inserting digit product with elements
    // in the vector pair
    for (var i = 0; i < n; i++) {
        vp[i].push(productOfDigit(arr[i]));
        vp[i].push(arr[i]);
    }
   
    // Sort the vector, this will sort the pair
    // according to the product of the digits
    vp.sort();
   
    // Print the sorted vector content
    for (var i = 0; i < n; i++)
        document.write(vp[i][1] + " ");
}
 
// Driver code
var arr = [ 12, 10, 102, 31, 15];
var n = arr.length;
sortArr(arr, n);
 
// This code is contributed by ShubhamSingh10
</script>

10 102 12 31 15

Complejidad de tiempo: O (nlogn)

Espacio Auxiliar: O(n)