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

  • Last Updated : 31 May, 2021

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++




// 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

 


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