Número mínimo de operaciones dadas requeridas para convertir n en m

Dados dos enteros n y m , en una sola operación n puede multiplicarse por 2 o por 3 . La tarea es convertir n en m con un número mínimo de operaciones dadas. Si es imposible convertir n en m con la operación dada, imprima -1 .
Ejemplos: 
 

Entrada: n = 120, m = 51840 
Salida: 7 
120 * 2 * 2 * 2 * 2 * 3 * 3 * 3 = 51840
Entrada: n = 42, m = 42 
Salida: 0 
No requiere operación.
Entrada: n = 48, m = 72 
Salida: -1 
 

Enfoque: si m no es divisible por n , imprima -1 ya que n no se puede convertir en m con la operación dada. De lo contrario, podemos comprobar si, al dividir, el cociente tiene solo 2 y 3 como factores primos. En caso afirmativo, el resultado será la suma de las potencias de 2 y 3 ; de lo contrario, imprima -1
. 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 minimum
// operations required
int minOperations(int n, int m)
{
    if (m % n != 0)
        return -1;
 
    int minOperations = 0;
    int q = m / n;
 
    // Counting all 2s
    while (q % 2 == 0) {
        q = q / 2;
        minOperations++;
    }
 
    // Counting all 3s
    while (q % 3 == 0) {
        q = q / 3;
        minOperations++;
    }
 
    // If q contained only 2 and 3
    // as the only prime factors
    // then it must be 1 now
    if (q == 1)
        return minOperations;
 
    return -1;
}
 
// Driver code
int main()
{
    int n = 120, m = 51840;
    cout << minOperations(n, m);
 
    return 0;
}

// Java implementation of the approach
class GfG {
 
    // Function to return the minimum
    // operations required
    static int minOperations(int n, int m)
    {
        if (m % n != 0)
            return -1;
 
        int minOperations = 0;
        int q = m / n;
 
        // Counting all 2s
        while (q % 2 == 0) {
            q = q / 2;
            minOperations++;
        }
 
        // Counting all 3s
        while (q % 3 == 0) {
            q = q / 3;
            minOperations++;
        }
 
        // If q contained only 2 and 3
        // as the only prime factors
        // then it must be 1 now
        if (q == 1)
            return minOperations;
 
        return -1;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 120, m = 51840;
        System.out.println(minOperations(n, m));
    }
}

# Python 3 implementation of the approach
 
# Function to return the minimum
# operations required
def minOperations(n, m):
    if (m % n != 0):
        return -1
 
    minOperations = 0
    q = int(m / n)
 
    # Counting all 2s
    while (q % 2 == 0):
        q = int(q / 2)
        minOperations += 1
 
    # Counting all 3s
    while (q % 3 == 0):
        q = int(q / 3)
        minOperations += 1
 
    # If q contained only 2 and 3
    # as the only prime factors
    # then it must be 1 now
    if (q == 1):
        return minOperations
 
    return -1
 
# Driver code
if __name__ == '__main__':
    n = 120
    m = 51840
    print(minOperations(n, m))
     
# This code is contributed by
# Surendra_Gangwar

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the minimum
// operations required
static int minOperations(int n, int m)
{
    if (m % n != 0)
        return -1;
 
    int minOperations = 0;
    int q = m / n;
 
    // Counting all 2s
    while (q % 2 == 0)
    {
        q = q / 2;
        minOperations++;
    }
 
    // Counting all 3s
    while (q % 3 == 0)
    {
        q = q / 3;
        minOperations++;
    }
 
    // If q contained only 2 and 3
    // as the only prime factors
    // then it must be 1 now
    if (q == 1)
        return minOperations;
 
    return -1;
}
 
// Driver code
public static void Main()
{
    int n = 120, m = 51840;
    Console.WriteLine(minOperations(n, m));
}
}
 
// This code is contributed
// by Akanksha Rai

<?php
// PHP implementation of the approach
 
// Function to return the minimum
// operations required
function minOperations($n, $m)
{
    if ($m % $n != 0)
        return -1;
 
    $minOperations = 0;
    $q = $m / $n;
 
    // Counting all 2s
    while ($q % 2 == 0)
    {
        $q = $q / 2;
        $minOperations++;
    }
 
    // Counting all 3s
    while ($q % 3 == 0)
    {
        $q = $q / 3;
        $minOperations++;
    }
 
    // If q contained only 2 and 3
    // as the only prime factors
    // then it must be 1 now
    if ($q == 1)
        return $minOperations;
 
    return -1;
}
 
// Driver code
$n = 120; $m = 51840;
echo(minOperations($n, $m));
 
// This code is contributed by Code_Mech
?>

<script>
// javascript implementation of the approach
 
    // Function to return the minimum
    // operations required
    function minOperations(n , m) {
        if (m % n != 0)
            return -1;
 
        var minOperations = 0;
        var q = m / n;
 
        // Counting all 2s
        while (q % 2 == 0) {
            q = q / 2;
            minOperations++;
        }
 
        // Counting all 3s
        while (q % 3 == 0) {
            q = q / 3;
            minOperations++;
        }
 
        // If q contained only 2 and 3
        // as the only prime factors
        // then it must be 1 now
        if (q == 1)
            return minOperations;
 
        return -1;
    }
 
    // Driver code
     
        var n = 120, m = 51840;
        document.write(minOperations(n, m));
 
// This code contributed by Rajput-Ji
</script>

7

Complejidad del tiempo: O(log(m/n))

Espacio Auxiliar: O(1)