Reduce a number to 1 by performing given operations

Given a number N. The task is to reduce the given number N to 1 in the minimum number of steps. You can perform any one of the below operations in each step.

  • Operation 1: If the number is even then you can divide the number by 2.
  • Operation 2: If the number is odd then you are allowed to perform either (n+1) or (n-1).

You need to print the minimum number of steps required to reduce the number N to 1 by performing the above operations.

Examples:



Input : n = 15
Output : 5
 15 is odd 15+1=16    
 16 is even 16/2=8     
 8  is even 8/2=4 
 4  is even 4/2=2     
 2  is even 2/2=1     

Input : n = 7
Output : 4
    7->6    
    6->3 
    3->2    
    2->1

The idea is to recursively compute the minimum number of steps required.

  • If the number is even, then we are allowed to only divide the number by 2.
  • But, when the number is Odd, we can either increment or decrement it by 1. So, we will use recursion for both n-1 and n+1 and return the one with the minimum number of operations.

Below is the implementation of above approach:

C++

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// CPP program to count minimum 
// steps to reduce a number
#include <iostream>
#include <cmath>
  
using namespace std;
  
int countways(int n)
{
    if (n == 1)
        return 0;
    else if (n % 2 == 0)
        return 1 + countways(n / 2);
    else
        return 1 + min(countways(n - 1), 
                       countways(n + 1));
}
  
// Driver code
int main()
{
    int n = 15;
  
    cout << countways(n)<< "\n";
  
    return 0;
}

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Java














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// Java program to count minimum  


// steps to reduce a number 


class Geeks { 


  


static int countways(int n) 


{ 


    if (n == 1) 


        return 0; 


    else if (n % 2 == 0) 


        return 1 + countways(n / 2); 


    else


        return 1 + Math.min(countways(n - 1), 


                            countways(n + 1)); 


} 


  


// Driver code 


public static void main(String args[]) 


{ 


    int n = 15; 


  


    System.out.println(countways(n)); 


  


} 


} 


  


// This code is contributed by ankita_saini 



















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Python3














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# Python program to count minimum  


# steps to reduce a number 


  


  


def countways(n): 


    if (n == 1): 


        return 0; 


    elif (n % 2 == 0): 


        return 1 + countways(n / 2); 


    else: 


        return 1 + min(countways(n - 1),  


                    countways(n + 1)); 


  


# Driver code 


n = 15; 


print(countways(n)); 


  


# This code is contributed by PrinciRaj1992 



















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C#














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// C# program to count minimum  


// steps to reduce a number 


using System; 


  


class GFG  


{ 


static int countways(int n) 


{ 


    if (n == 1) 


        return 0; 


    else if (n % 2 == 0) 


        return 1 + countways(n / 2); 


    else


        return 1 + Math.Min(countways(n - 1), 


                            countways(n + 1)); 


} 


  


// Driver code 


static public void Main () 


{ 


    int n = 15; 


    Console.Write(countways(n)); 


} 


} 


  


// This code is contributed by Raj 



















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Output:


5






          
          
          

          

    

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