Minimum multiplications with {2, 3, 7} to make two numbers equal

Given two numbers A and B, the task is to find the minimum number of operations required to make A and B equal. In each operation, any number can be divided by either 2, 3 or 7. If it is not possible then print -1.

Examples:

Input: A = 14, B = 28
Output: 1
Operation 1: A * 2 = 14 * 2 = 28 which is equal to B.

Input: A = 3, B = 5
Output: -1
No matter how many times the operation is performed, A and B will never be equal.



Approach: A and B can be written as A = x * 2a1 * 3a2 * 7a3 and B = y * 2b1 * 3b2 * 7b3 where x and y are not divisible by 2, 3 and 7. Now,

  • If x != y then A and B cannot be made equal with the given operation.
  • If x = y then the minimum operations required will be |a1 – b1| + |a2 – b2| + |a3 – b3| because both the numbers need to have equal powers of 2, 3 and 7.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find powers of 2, 3 and 7 in x
vector<int> Divisors(int x)
{
    // To keep count of each divisor
    int c = 0;
  
    // To store the result
    vector<int> v;
  
    // Count powers of 2 in x
    while (x % 2 == 0) {
        c++;
        x /= 2;
    }
    v.push_back(c);
  
    c = 0;
  
    // Count powers of 3 in x
    while (x % 3 == 0) {
        c++;
        x /= 3;
    }
    v.push_back(c);
  
    c = 0;
  
    // Count powers of 7 in x
    while (x % 7 == 0) {
        c++;
        x /= 7;
    }
    v.push_back(c);
  
    // Reamining number which is not
    // divisible by 2, 3 or 7
    v.push_back(x);
  
    return v;
}
  
// Function to return the minimum number of
// given operations required to make a and b equal
int MinOperations(int a, int b)
{
    // a = x * 2^a1 * 3^a2 * 7^a3
    // va[0] = a1
    // va[1] = a2
    // va[2] = a3
    // va[3] = x
    vector<int> va = Divisors(a);
  
    // Similarly for b
    vector<int> vb = Divisors(b);
  
    // If a and b cannot be made equal
    // with the given operation. Note
    // that va[3] and vb[3] contain
    // remaining numbers after repeated 
    // divisions with 2, 3 and 7.
    // If remaining numbers are not same
    // then we cannot make them equal.
    if (va[3] != vb[3])
        return -1;
  
    // Minimum number of operations required
    int minOperations = abs(va[0] - vb[0])
                        + abs(va[1] - vb[1])
                        + abs(va[2] - vb[2]);
  
    return minOperations;
}
  
// Driver code
int main()
{
    int a = 14, b = 28;
    cout << MinOperations(a, b);
  
    return 0;
}

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Python3














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# python 3 implementation of the approach 


  


# Function to find powers of 2, 3 and 7 in x 


def Divisors(x): 


    # To keep count of each divisor 


    c = 0


  


    # To store the result 


    v = [] 


  


    # Count powers of 2 in x 


    while (x % 2 == 0): 


        c += 1


        x /= 2


  


    v.append(c) 


  


    c = 0


  


    # Count powers of 3 in x 


    while (x % 3 == 0): 


        c += 1


        x /= 3


  


    v.append(c) 


  


    c = 0


  


    # Count powers of 7 in x 


    while (x % 7 == 0): 


        c += 1


        x /= 7


  


    v.append(c) 


  


    # Reamining number which is not 


    # divisible by 2, 3 or 7 


    v.append(x) 


  


    return v 


  


# Function to return the minimum number of 


# given operations required to make a and b equal 


def MinOperations(a, b): 


    # a = x * 2^a1 * 3^a2 * 7^a3 


    # va[0] = a1 


    # va[1] = a2 


    # va[2] = a3 


    # va[3] = x 


    va = Divisors(a) 


  


    # Similarly for b 


    vb = Divisors(b) 


  


    # If a and b cannot be made equal 


    # with the given operation. Note 


    # that va[3] and vb[3] contain 


    # remaining numbers after repeated  


    # divisions with 2, 3 and 7. 


    # If remaining numbers are not same 


    # then we cannot make them equal. 


    if (va[3] != vb[3]): 


        return -1


  


    # Minimum number of operations required 


    minOperations = abs(va[0] - vb[0]) + abs(va[1] - vb[1]) + abs(va[2] - vb[2]) 


  


    return minOperations 


  


# Driver code 


if __name__ == '__main__': 


    a = 14


    b = 28


    print(MinOperations(a, b)) 


  


# This code is contributed by 


# Sanjit_Prasad 



















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PHP














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<?php 


// PHP implementation of the approach  


  


// Function to find powers of 2, 3 and 7 in x  


function Divisors($x)  


{  


    // To keep count of each divisor  


    $c = 0;  


  


    // To store the result  


    $v = array();  


  


    // Count powers of 2 in x  


    while ($x % 2 == 0)  


    {  


        $c++;  


        $x = floor($x / 2);  


    }  


    array_push($v, $c); 


  


    $c = 0;  


  


    // Count powers of 3 in x  


    while ($x % 3 == 0) 


    {  


        $c++;  


        $x = floor($x / 3);  


    }  


    array_push($v, $c) ; 


  


    $c = 0;  


  


    // Count powers of 7 in x  


    while ($x % 7 == 0)  


    {  


        $c++;  


        $x = floor($x / 7);  


    }  


    array_push($v, $c);  


  


    // Reamining number which is not  


    // divisible by 2, 3 or 7  


    array_push($v, $x);  


  


    return $v;  


}  


  


// Function to return the minimum number  


// of given operations required to make  


// a and b equal  


function MinOperations($a, $b)  


{  


      


    // a = x * 2^a1 * 3^a2 * 7^a3  


    // va[0] = a1  


    // va[1] = a2  


    // va[2] = a3  


    // va[3] = x  


    $va = Divisors($a);  


  


    // Similarly for b  


    $vb = Divisors($b);  


  


    // If a and b cannot be made equal  


    // with the given operation. Note  


    // that va[3] and vb[3] contain  


    // remaining numbers after repeated  


    // divisions with 2, 3 and 7.  


    // If remaining numbers are not same  


    // then we cannot make them equal.  


    if ($va[3] != $vb[3])  


        return -1;  


  


    // Minimum number of operations required  


    $minOperations = abs($va[0] - $vb[0]) +  


                     abs($va[1] - $vb[1]) +  


                     abs($va[2] - $vb[2]);  


  


    return $minOperations;  


}  


  


// Driver code  


$a = 14 ; 


$b = 28 ; 


echo MinOperations($a, $b);  


  


// This code is contributed by Ryuga 


?> 



















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