# Convert a number m to n using minimum number of given operations

• Difficulty Level : Medium
• Last Updated : 01 Jun, 2022

Convert a number m to n with minimum operations. The operations allowed are :

1. Multiply by 2, i.e., do m = 2 * m
2. Subtract 1, i.e., do m = m – 1

Print -1 if it is not possible to convert.
Examples :

```Input : m = 3, n = 11
Output : 3
1st operation: *2 = 3*2 = 6
2nd operation: *2 = 6*2 = 12
3rd operation: -1 = 12-1 = 11

Input : m = 15, n = 20
Output : 6
1st operation: -1 '5' times = 15 + (-1*5) = 10
2nd operation: *2 = 10*2 = 20

Input : m = 0, n = 8
Output : -1
Using the given set of operations 'm'
cannot be converted to 'n'

Input : m = 12, n = 8
Output : 4```

The idea is based on below facts.
1) If m is less than 0 and n is greater than 0, then not possible.
2) If m is greater than n, then we can reach n using subtractions only.
3) Else (m is less than n), we must do m*2 operations. Following two cases arise.
……a) If n is odd, we must do a -1 operation to reach it.
……b) If n is even, we must do a *2 operation to reach it.
Algorithm:

```int convert(m,n)
if (m == n)
return 0;

// not possible
if (m <= 0 && n > 0)
return -1;

// m is greater than n
if (m > n)
return m-n;

// n is odd
if (n % 2 == 1)
// perform '-1'
return 1 + convert(m, n+1);

// n is even
else
// perform '*2'
return 1 + convert(m, n/2);```

Note: The list of operations so generated should be applied in reverse order.
For example:

```m = 3, n = 11
convert(3,11)
|       --> n is odd:   operation '-1'
convert(3,12)
|       --> n is even:  operation '*2'
convert(3,6)
|       --> n is even:  operation '*2'
convert(3,3)
|       --> m == n
return
Therefore, the sequence of operations is '*2', '*2', '-1'.```

## C++

 `// C++ implementation to convert``// a number m to n using minimum``// number of given operations``#include ``using` `namespace` `std;` `// Function to find minimum``// number of given operations``// to convert m to n``int` `convert(``int` `m, ``int` `n)``{``    ``if` `(m == n)``        ``return` `0;` `    ``// only way is to do``    ``// -1 (m - n) times``    ``if` `(m > n)``        ``return` `m - n;` `    ``// not possible``    ``if` `(m <= 0 && n > 0)``        ``return` `-1;` `    ``// n is greater and n is odd``    ``if` `(n % 2 == 1)` `        ``// perform '-1' on m``        ``// (or +1 on n)``        ``return` `1 + convert(m, n + 1);` `    ``// n is even``    ``else``        ``// perform '*2' on m``        ``// (or n/2 on n)``        ``return` `1 + convert(m, n / 2);``}` `// Driver code``int` `main()``{``    ``int` `m = 3, n = 11;``    ``cout << ``"Minimum number of operations : "``         ``<< convert(m, n);``    ``return` `0;``}`

## Java

 `// Java implementation to convert``// a number m to n using minimum``// number of given operations` `class` `ConvertNum``{``    ``// function to find minimum``    ``// number of given operations``    ``// to convert m to n``    ``static` `int` `convert(``int` `m, ``int` `n)``    ``{``        ``if` `(m == n)``            ``return` `0``;``    ` `        ``// only way is to do``        ``// -1 (m - n) times``        ``if` `(m > n)``            ``return` `m - n;``    ` `        ``// not possible``        ``if` `(m <= ``0` `&& n > ``0``)``            ``return` `-``1``;``    ` `        ``// n is greater and n is odd``        ``if` `(n % ``2` `== ``1``)``    ` `            ``// perform '-1' on m``            ``// (or +1 on n)``            ``return` `1` `+ convert(m, n + ``1``);``    ` `        ``// n is even``        ``else``            ``// perform '*2' on m``            ``// (or n / 2 on n)``            ``return` `1` `+ convert(m, n / ``2``);``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `m = ``3``, n = ``11``;``        ``System.out.println(``"Minimum number of "` `+``                                ``"operations : "` `+``                                  ``convert(m, n));``    ``}``}`

## Python3

 `# Python implementation to convert``# a number m to n using minimum``# number of given operations` `# Function to find minimum``# number of given operations``# to convert m to n``def` `convert(m, n):` `    ``if``(m ``=``=` `n):``        ``return` `0` `    ``# only way is to do``    ``# -1(m - n): times``    ``if``(m > n):``        ``return` `m ``-` `n` `    ``# not possible``    ``if``(m <``=` `0` `and` `n > ``0``):``        ``return` `-``1` `    ``# n is greater and n is odd``    ``if``(n ``%` `2` `=``=` `1``):` `        ``# perform '-1' on m``        ``#(or +1 on n):``        ``return` `1` `+` `convert(m, n ``+` `1``)` `    ``# n is even``    ``else``:``        ` `        ``# perform '*2' on m``        ``#(or n/2 on n):``        ``return` `1` `+` `convert(m, n ``/` `2``)` `# Driver code``m ``=` `3``n ``=` `11``print``(``"Minimum number of operations :"``,``                          ``convert(m, n))` `# This code is contributed by``# Sanjit_Prasad`

## C#

 `// C# implementation to convert``// a number m to n using minimum``// number of given operations``using` `System;` `class` `GFG``{``    ``// function to find minimum``    ``// number of given operations``    ``// to convert m to n``    ``static` `int` `convert(``int` `m, ``int` `n)``    ``{``        ``if` `(m == n)``            ``return` `0;``    ` `        ``// only way is to do``        ``// -1 (m - n) times``        ``if` `(m > n)``            ``return` `m - n;``    ` `        ``// not possible``        ``if` `(m <= 0 && n > 0)``            ``return` `-1;``    ` `        ``// n is greater and n is odd``        ``if` `(n % 2 == 1)``    ` `            ``// perform '-1' on m``            ``// (or +1 on n)``            ``return` `1 + convert(m, n + 1);``    ` `        ``// n is even``        ``else``            ``// perform '*2' on m``            ``// (or n / 2 on n)``            ``return` `1 + convert(m, n / 2);``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `m = 3, n = 11;``        ``Console.Write(``"Minimum number of "` `+``                           ``"operations : "` `+``                             ``convert(m, n));``    ``}``}` `// This code is contributed by nitin mittal`

## PHP

 ` ``\$n``)``        ``return` `\$m` `- ``\$n``;` `    ``// not possible``    ``if` `(``\$m` `<= 0 && ``\$n` `> 0)``        ``return` `-1;` `    ``// n is greater and n is odd``    ``if` `(``\$n` `% 2 == 1)` `        ``// perform '-1' on m``        ``// (or +1 on n)``        ``return` `1 + convert(``\$m``, ``\$n` `+ 1);` `    ``// n is even``    ``else``        ``// perform '*2' on m``        ``// (or n/2 on n)``        ``return` `1 + convert(``\$m``, ``\$n` `/ 2);``}` `// Driver code``{``    ``\$m` `= 3; ``\$n` `= 11;``    ``echo` `"Minimum number of "``.``              ``"operations : "``,``              ``convert(``\$m``, ``\$n``);``    ``return` `0;``}    ` `// This code is contributed``// by nitin mittal.``?>`

## Javascript

 ``

Output :

`Minimum number of operations : 3`

References :
http://tech.queryhome.com/112705/convert-number-with-minimum-operations-operations-allowed
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