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

**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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` |

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## 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 ` ` ` `?> ` |

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**Output:**

5

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