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

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

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**.

**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:

**N is power of 2 :**reduce n to n/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.

**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.**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 <iostream>` `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

`<?php` `// PHP program to find the number of step to reduce n to 1` `// recursive function to` `// count set bits` `function` `countSetBits(` `$n` `)` `{` ` ` `// base case` ` ` `if` `(` `$n` `== 0)` ` ` `return` `0;` ` ` `else` ` ` `return` `1 + ` ` ` `countSetBits(` `$n` `& ` ` ` `(` `$n` `- 1));` `}` `// Function to return number of steps for reduction` `function` `stepRequired(` `$n` `)` `{` ` ` ` ` `return` `countSetBits(` `$n` `- 1);` `}` ` ` `// Driver program` `$n` `= 94;` `echo` `stepRequired(` `$n` `);` `// This code is contributed by` `// ihritik` `?>` |

## Javascript

`<script>` ` ` `// Javascript program to find the number of step to reduce n to 1` ` ` ` ` `// function to count set bits` ` ` `function` `countSetBits(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` ` ` `function` `stepRequired(n)` ` ` `{` ` ` ` ` `return` `countSetBits(n - 1);` ` ` `}` ` ` ` ` `let n = 94;` ` ` `document.write(stepRequired(n));` `// This code is contributed by decode2207.` `</script>` |

**Output:**

5