Skip to content
Related Articles
Minimum number operations required to convert n to m | Set-2
• Last Updated : 23 Apr, 2021

Given two integers n and m and a and b, in a single operation n can be multiplied by either a or b. The task is to convert n to m with a minimum number of given operations. If it is impossible to convert n to m with the given operation then print -1.

Examples:

```Input: n = 120, m = 51840, a = 2, b = 3
Output: 7
120 * 2 * 2 * 2 * 2 * 3 * 3 * 3 = 51840

Input: n = 10, m = 50, a = 5, b = 7
Output: 1
10 * 5 = 50 ```

In the previous post, we discussed an approach using division.
In this post, we will use an approach that finds the minimum number of operations using recursion. The recursion will consist of two states, the number being multiplied by a or by b, and counting the steps. The minimum of both steps will be the answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach` `#include ``using` `namespace` `std;``#define MAXN 10000000` `// Function to find the minimum number of steps``int` `minimumSteps(``int` `n, ``int` `m, ``int` `a, ``int` `b)``{``    ``// If n exceeds M``    ``if` `(n > m)``        ``return` `MAXN;` `    ``// If N reaches the target``    ``if` `(n == m)``        ``return` `0;` `    ``// The minimum of both the states``    ``// will be the answer``    ``return` `min(1 + minimumSteps(n * a, m, a, b),``               ``1 + minimumSteps(n * b, m, a, b));``}` `// Driver code``int` `main()``{``    ``int` `n = 120, m = 51840;``    ``int` `a = 2, b = 3;``    ``cout << minimumSteps(n, m, a, b);``    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``class` `GFG``{``    ``static` `int` `MAXN = ``10000000``;``    ` `    ``// Function to find the minimum number of steps``    ``static` `int` `minimumSteps(``int` `n, ``int` `m, ``int` `a, ``int` `b)``    ``{``        ``// If n exceeds M``        ``if` `(n > m)``            ``return` `MAXN;``    ` `        ``// If N reaches the target``        ``if` `(n == m)``            ``return` `0``;``    ` `        ``// The minimum of both the states``        ``// will be the answer``        ``return` `Math.min(``1` `+ minimumSteps(n * a, m, a, b),``                ``1` `+ minimumSteps(n * b, m, a, b));``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `n = ``120``, m = ``51840``;``        ``int` `a = ``2``, b = ``3``;``        ``System.out.println(minimumSteps(n, m, a, b));``    ``}``}` `// This code is contributed by ihritik`

## Python3

 `# Python 3 implementation of the``# above approach``MAXN ``=` `10000000` `# Function to find the minimum``# number of steps``def` `minimumSteps(n, m, a, b):``    ` `    ``# If n exceeds M``    ``if` `(n > m):``        ``return` `MAXN` `    ``# If N reaches the target``    ``if` `(n ``=``=` `m):``        ``return` `0` `    ``# The minimum of both the states``    ``# will be the answer``    ``return` `min``(``1` `+` `minimumSteps(n ``*` `a, m, a, b),``               ``1` `+` `minimumSteps(n ``*` `b, m, a, b))` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``n ``=` `120``    ``m ``=` `51840``    ``a ``=` `2``    ``b ``=` `3``    ``print``(minimumSteps(n, m, a, b))` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# implementation of the above approach``using` `System;` `class` `GFG``{``    ``static` `int` `MAXN = 10000000;``    ` `    ``// Function to find the minimum number of steps``    ``static` `int` `minimumSteps(``int` `n, ``int` `m, ``int` `a, ``int` `b)``    ``{``        ``// If n exceeds M``        ``if` `(n > m)``            ``return` `MAXN;``    ` `        ``// If N reaches the target``        ``if` `(n == m)``            ``return` `0;``    ` `        ``// The minimum of both the states``        ``// will be the answer``        ``return` `Math.Min(1 + minimumSteps(n * a, m, a, b),``                ``1 + minimumSteps(n * b, m, a, b));``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `n = 120, m = 51840;``        ``int` `a = 2, b = 3;``        ``Console.WriteLine(minimumSteps(n, m, a, b));``    ``}``}` `// This code is contributed by ihritik`

## PHP

 ` ``\$m``)``        ``return` `\$MAXN``;` `    ``// If N reaches the target``    ``if` `(``\$n` `== ``\$m``)``        ``return` `0;` `    ``// The minimum of both the states``    ``// will be the answer``    ``return` `min(1 + minimumSteps(``\$n` `* ``\$a``, ``\$m``, ``\$a``, ``\$b``),``               ``1 + minimumSteps(``\$n` `* ``\$b``, ``\$m``, ``\$a``, ``\$b``));``}` `// Driver code``\$n` `= 120; ``\$m` `= 51840;``\$a` `= 2; ``\$b` `= 3;``echo` `minimumSteps(``\$n``, ``\$m``, ``\$a``, ``\$b``);` `// This code is contributed by Akanksha Rai``?>`

## Javascript

 ``
Output:
`7`

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer DSA Live Classes

My Personal Notes arrow_drop_up