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

Given a positive integer N    . 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: 
 

// 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 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 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# 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 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

?>

<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>

5

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

Auxiliary Space: O(1)