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

• Last Updated : 31 May, 2021

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`

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