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:

C++

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// 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

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// 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

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# 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#

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// 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

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<?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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Improved By : AnkitRai01, ihritik