# Count number of steps to cover a distance if steps can be taken in powers of 2

• Difficulty Level : Medium
• Last Updated : 10 Nov, 2021

Given a distance K to cover, the task is to find the minimum steps required to cover the distance if steps can be taken in powers of 2 like 1, 2, 4, 8, 16……..
Examples :

```Input : K = 9
Output : 2

Input : K = 343
Output : 6```

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.

The minimum steps required can be calculated by reducing K by the highest power of 2 in each step which can be obtained by counting no. of set bits in the binary representation of a number.
Below is the implementation of the above approach:

## C++

 `// C++ program to count the minimum number of steps` `#include ``using` `namespace` `std;` `// Function to count the minimum number of steps``int` `getMinSteps(``int` `K)``{``   ``// __builtin_popcount() is a C++ function to``   ``// count the number of set bits in a number``   ``return` `__builtin_popcount(k);``}` `// Driver Code``int` `main()``{``    ``int` `n = 343;``    ` `    ``cout << getMinSteps(n)<< ``'\n'``;` `    ``return` `0;``}`

## Java

 `// Java program to count minimum number of steps``import` `java.io.*;` `class` `GFG``{``    ` `    ``// Function to count the minimum number of steps``    ``static` `int` `getMinSteps(``int` `K)``    ``{``        ``// count the number of set bits in a number``        ``return` `Integer.bitCount(K);``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `n = ``343``;``        ` `        ``System.out.println(getMinSteps(n));``    ``}``}` `// This code is contributed by AnkitRai01`

## Python3

 `# Python 3 implementation of the approach` `# Function to count the minimum number of steps``def` `getMinSteps(K) :``    ` `    ``# bin(K).count("1") is a Python3 function to``    ``# count the number of set bits in a number``    ``return` `bin``(K).count(``"1"``)` `# Driver Code``n ``=` `343``print``(getMinSteps(n))` `# This code is contributed by``# divyamohan123`

## C#

 `// C# program to count minimum number of steps``using` `System;``    ` `class` `GFG``{``    ` `    ``// Function to count the minimum number of steps``    ``static` `int` `getMinSteps(``int` `K)``    ``{``        ``// count the number of set bits in a number``        ``return` `countSetBits(K);``    ``}``    ` `    ``static` `int` `countSetBits(``int` `x)``    ``{``        ``int` `setBits = 0;``        ``while` `(x != 0)``        ``{``            ``x = x & (x - 1);``            ``setBits++;``        ``}``        ``return` `setBits;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main (String[] args)``    ``{``        ``int` `n = 343;``        ` `        ``Console.WriteLine(getMinSteps(n));``    ``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output:
`6`

Time Complexity :

Auxiliary Space: O(1)

My Personal Notes arrow_drop_up