Find the minimum number of steps to reach M from N

• Difficulty Level : Easy
• Last Updated : 07 Apr, 2021

Given two integers N and M. The task is to find the minimum number of steps to reach M from N by performing given operations.

1. Multiply a number x by 2. So, x becomes 2*x.
2. Subtract one from the number x. So, x becomes x-1.

Examples:

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 experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

```Input : N = 4, M = 6
Output : 2
Explanation : Perform operation number 2 on N.
So, N becomes 3 and then perform operation number 1.
Then, N becomes 6. So, the minimum number of steps is 2.

Input : N = 10, M = 1
Output : 9
Explanation : Perform operation number two
9 times on N. Then N becomes 1.```

Approach
The idea is to reverse the problem as follows: We should get the number N starting from M using the operations:

1. Divide the number by 2 if it is even.
2. Add 1 to the number.

Now, the minimum number of operations would be:

1. If N > M, return the difference between them, that is, number of steps will be adding 1 to M until it becomes equal to N.
2. Else if N < M.
• Keep dividing M by 2 until it becomes less than N. If M is odd, add 1 to it first and then divide by 2. Once M is less than N, add the difference between them to the count along with the count of above operations.

Below is the implementation of the above approach:

C++

 `// CPP program to find minimum number``// of steps to reach M from N``#include ``using` `namespace` `std;` `// Function to find a minimum number``// of steps to reach M from N``int` `Minsteps(``int` `n, ``int` `m)``{``    ``int` `ans = 0;``    ` `    ``// Continue till m is greater than n``    ``while``(m > n)``    ``{``        ``// If m is odd``        ``if``(m&1)``        ``{``            ``// add one``            ``m++;``            ``ans++;``        ``}``        ` `        ``// divide m by 2       ``        ``m /= 2;``        ``ans++;``    ``}``    ` `    ``// Return the required answer``    ``return` `ans + n - m;``}` `// Driver code``int` `main()``{``    ``int` `n = 4, m = 6;``   ` `    ``cout << Minsteps(n, m);``    ` `    ``return` `0;``}`

Java

 `// Java program to find minimum number``// of steps to reach M from N``class` `CFG``{``    ` `// Function to find a minimum number``// of steps to reach M from N``static` `int` `Minsteps(``int` `n, ``int` `m)``{``    ``int` `ans = ``0``;``    ` `    ``// Continue till m is greater than n``    ``while``(m > n)``    ``{``        ``// If m is odd``        ``if``(m % ``2` `!= ``0``)``        ``{``            ``// add one``            ``m++;``            ``ans++;``        ``}``        ` `        ``// divide m by 2    ``        ``m /= ``2``;``        ``ans++;``    ``}``    ` `    ``// Return the required answer``    ``return` `ans + n - m;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``4``, m = ``6``;``    ` `    ``System.out.println(Minsteps(n, m));``}``}` `// This code is contributed by Code_Mech`

Python3

 `# Python3 program to find minimum number``# of steps to reach M from N` `# Function to find a minimum number``# of steps to reach M from N``def` `Minsteps(n, m):` `    ``ans ``=` `0``    ` `    ``# Continue till m is greater than n``    ``while``(m > n):` `        ``# If m is odd``        ``if``(m & ``1``):``            ` `            ``# add one``            ``m ``+``=` `1``            ``ans ``+``=` `1``        ` `        ``# divide m by 2    ``        ``m ``/``/``=` `2``        ``ans ``+``=` `1``    ` `    ``# Return the required answer``    ``return` `ans ``+` `n ``-` `m` `# Driver code``n ``=` `4``m ``=` `6` `print``(Minsteps(n, m))` `# This code is contributed by mohit kumar`

C#

 `// C# program to find minimum number``// of steps to reach M from N``using` `System;` `class` `GFG``{``    ` `// Function to find a minimum number``// of steps to reach M from N``static` `int` `Minsteps(``int` `n, ``int` `m)``{``    ``int` `ans = 0;``    ` `    ``// Continue till m is greater than n``    ``while``(m > n)``    ``{``        ``// If m is odd``        ``if``(m % 2 != 0)``        ``{``            ``// add one``            ``m++;``            ``ans++;``        ``}``        ` `        ``// divide m by 2    ``        ``m /= 2;``        ``ans++;``    ``}``    ` `    ``// Return the required answer``    ``return` `ans + n - m;``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `n = 4, m = 6;``    ` `    ``Console.WriteLine(Minsteps(n, m));``}``}` `// This code is contributed``// by Akanksha Rai`

PHP

 ` ``\$n``)``    ``{``        ``// If m is odd``        ``if``(``\$m` `% 2 != 0)``        ``{``            ``// add one``            ``\$m``++;``            ``\$ans``++;``        ``}``        ` `        ``// divide m by 2    ``        ``\$m` `/= 2;``        ``\$ans``++;``    ``}``    ` `    ``// Return the required answer``    ``return` `\$ans` `+ ``\$n` `- ``\$m``;``}` `// Driver code``\$n` `= 4; ``\$m` `= 6;` `echo``(Minsteps(``\$n``, ``\$m``));` `// This code is contributed by Code_Mech``?>`

Javascript

 ``
Output:
`2`

My Personal Notes arrow_drop_up