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# Count number of step required to reduce N to 1 by following certain rule

• Last Updated : 30 Aug, 2022

Given a positive integer . Find the number of steps required to minimize it to 1. In a single step N either got reduced to half if it is power of 2 else N is reduced to difference of N and its nearest power of 2 which is smaller than N.

Examples:

```Input : N = 2
Output : 1```
```Input : N = 20
Output : 3```

Simple Approach: As per question a very simple and brute force approach is to iterate over N until it got reduced to 1, where reduction involve two cases:

1. N is power of 2 : reduce n to n/2
2. N is not power of 2: reduce n to n – (2^log2(n))

Efficient approach: Before proceeding to actual result lets have a look over bit representation of an integer n as per problem statement.

1. When an integer is power of 2: In this case bit -representation includes only one set bit and that too is left most. Hence log2(n) i.e. bit-position minus One is the number of step required to reduce it to n. Which is also equal to number of set bit in n-1.
2. When an integer is not power of 2:The remainder of n – 2^(log2(n)) is equal to integer which can be obtained by un-setting the left most set bit. Hence, one set bit removal count as one step in this case.

Hence the actual answer for steps required to reduce n is equal to number of set bits in n-1. Which can be easily calculated either by using the loop or any of method described in the post: Count Set bits in an Integer.

Below is the implementation of the above approach:

## C++

 `// Cpp to find the number of step to reduce n to 1``// C++ program to demonstrate __builtin_popcount()``#include ``using` `namespace` `std;` `// Function to return number of steps for reduction``int` `stepRequired(``int` `n)``{``    ``// builtin function to count set bits``    ``return` `__builtin_popcount(n - 1);``}` `// Driver program``int` `main()``{``    ``int` `n = 94;``    ``cout << stepRequired(n) << endl;``    ``return` `0;``}`

## Java

 `// Java program to find the number of step to reduce n to 1` `import` `java.io.*;``class` `GFG``{``    ``// Function to return number of steps for reduction``    ``static` `int` `stepRequired(``int` `n)``    ``{``        ``// builtin function to count set bits``        ``return` `Integer.bitCount(n - ``1``);``    ``}``    ` `    ``// Driver program``    ``public` `static` `void`  `main(String []args)``    ``{``        ``int` `n = ``94``;``        ``System.out.println(stepRequired(n));``    ` `    ``}``}`  `// This code is contributed by``// ihritik`

## Python3

 `# Python3 to find the number of step``# to reduce n to 1``# Python3 program to demonstrate``# __builtin_popcount()` `# Function to return number of``# steps for reduction``def` `stepRequired(n) :` `    ``# step to count set bits``    ``return` `bin``(``94``).count(``'1'``)` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:` `    ``n ``=` `94``    ``print``(stepRequired(n))` `# This code is contributed by Ryuga`

## C#

 `// C# program to find the number of step to reduce n to 1` `using` `System;``class` `GFG``{``    ` `    ``// function to count set bits``    ``static` `int` `countSetBits(``int` `n)``    ``{``  ` `        ``// base case``        ``if` `(n == 0)``            ``return` `0;``  ` `        ``else``  ` `            ``// if last bit set``            ``// add 1 else add 0``            ``return` `(n & 1) + countSetBits(n >> 1);``    ``}``    ``// Function to return number of steps for reduction``    ``static` `int` `stepRequired(``int` `n)``    ``{``     ` `        ``return` `countSetBits(n - 1);``    ``}``    ` `    ``// Driver program``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 94;``        ``Console.WriteLine(stepRequired(n));``    ` `    ``}``}`  `// This code is contributed by``// ihritik`

## PHP

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## Javascript

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Output:

`5`

Time Complexity: O(1) as constant time is taken.

Auxiliary Space: O(1)

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