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Convert a number m to n using minimum number of given operations
  • Difficulty Level : Medium
  • Last Updated : 05 Apr, 2021

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 <bits/stdc++.h>
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 conver(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 + conver(m, n + 1)
 
    # n is even
    else:
         
        # perform '*2' on m
        #(or n/2 on n):
        return 1 + conver(m, n / 2)
 
# Driver code
m = 3
n = 11
print("Minimum number of operations :",
                          conver(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



<?php
// PHP 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
function 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 && $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



<script>
// javascript 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
function 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 && 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
var m = 3, n = 11;
document.write("Minimum number of " +
                        "operations : " +
                          convert(m, n));
 
// This code is contributed by Princi Singh
</script>
Output : 
 
Minimum number of operations : 3
References : 
http://tech.queryhome.com/112705/convert-number-with-minimum-operations-operations-allowed
This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 
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