Reduce a number to 1 by performing given operations | Set 3

Given an integer N, the task is to find the number of steps required to reduce the given number N to 1 by performing the following operations:

1. If the number is a power of 2, then divide the number by 2.
2. Otherwise, subtract the greatest power of 2 smaller than N from N.

Examples:

Input: N = 2 
Output: 1
Explanation: The given number can be reduced to 1 by following the following steps:
Divide the number by 2 as N is a power of 2 which modifies the N to 1.
Therefore, the N can be reduced to 1 in only 1 step.

Input: N = 7 
Output: 2
Explanation: The given number can be reduced to 1 by following the following steps:
Subtract 4 the greatest power of 2 less than N. After the step the N modifies to N = 7-4 = 3.
Subtract 2 the greatest power of 2 less than N. After the step the N modifies to N = 3-2 = 1.
Therefore, the N can be reduced to 1 in 2 steps.

Method 1 –  

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

• If the number is a power of 2, then we are allowed to only divide the number by 2.
• Else we can subtract the greatest power of 2 smaller than N from N.
• So, we will use recursion for both the operations whenever valid and return number of operations.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Utility function to check
// if n is power of 2
int highestPowerof2(int n)
{
   int p = (int)log2(n);
   return (int)pow(2, p);
}

// Utility function to find highest
// power of 2 less than or equal
// to given number
bool isPowerOfTwo(int n)
{
   if(n==0)
   return false;

   return (ceil(log2(n)) == floor(log2(n)));
}

// Recursive function to find
// steps needed to reduce
// a given integer to 1
int reduceToOne(int N)
{
    // Base Condition
    if(N == 1){
        return 0;
    }
    // If the number is a power of 2
    if(isPowerOfTwo(N) == true){
        return 1 + reduceToOne(N/2);
    }
    // Else subtract the greatest
    //power of 2 smaller than N
    //from N
    else{
        return 1 + reduceToOne(N - highestPowerof2(N));
    }
}

// Driver Code
int main()
{
    // Input
    int N = 7;
    // Function call
    cout << reduceToOne(N) << endl;
}

Java

// java program for the above approach

class GFG {

  // Utility function to check
  // if n is power of 2
  static int highestPowerof2(int n)
  {
     int p = (int)(Math.log(n) / Math.log(2));
     return (int)Math.pow(2, p);
  }

  // Utility function to find highest
  // power of 2 less than or equal
  // to given number
  static boolean isPowerOfTwo(int n)
  {
     if(n==0)
     return false;

     return (int)(Math.ceil((Math.log(n) / Math.log(2)))) ==
       (int)(Math.floor(((Math.log(n) / Math.log(2)))));
  }

  // Recursive function to find
  // steps needed to reduce
  // a given integer to 1
  static int reduceToOne(int N)
  {
      // Base Condition
      if(N == 1){
          return 0;
      }
      // If the number is a power of 2
      if(isPowerOfTwo(N) == true){
          return 1 + reduceToOne(N/2);
      }
      // Else subtract the greatest
      //power of 2 smaller than N
      //from N
      else{
          return 1 + reduceToOne(N - highestPowerof2(N));
      }
  }

  // Driver Code
  public static void main(String [] args)
  {
      // Input
      int N = 7;
      // Function call
      System.out.println(reduceToOne(N));
  }

}


// This code is contributed by ihritik

Python

# Python program for the above approach
import math

# Utility function to check
# Log base 2
def Log2(x):
    if x == 0:
        return false;

    return (math.log10(x) /
            math.log10(2));

# Utility function to check
# if x is power of 2
def isPowerOfTwo(n):
    return (math.ceil(Log2(n)) ==
            math.floor(Log2(n)));

# Utility function to find highest
# power of 2 less than or equal
# to given number
def highestPowerof2(n):

    p = int(math.log(n, 2));
    return int(pow(2, p));

# Recursive function to find
# steps needed to reduce
# a given integer to 1
def reduceToOne(N):

    # Base Condition
    if(N == 1):
        return 0

    # If the number is a power of 2
    if(isPowerOfTwo(N) == True):
        return 1 + reduceToOne(N/2)

    # Else subtract the greatest
    # power of 2 smaller than N
    # from N
    else:
        return 1 + reduceToOne(N - highestPowerof2(N))

# Driver Code

# Input
N = 7;

#Function call
print(reduceToOne(N))

# This code is contributed by Samim Hossain Mondal.

C#

// C# program for the above approach
using System;

class GFG {

  // Utility function to check
  // if n is power of 2
  static int highestPowerof2(int n)
  {
     int p = (int)(Math.Log(n) / Math.Log(2));
     return (int)Math.Pow(2, p);
  }

  // Utility function to find highest
  // power of 2 less than or equal
  // to given number
  static bool isPowerOfTwo(int n)
  {
     if(n == 0)
     return false;

     return (int)(Math.Ceiling((Math.Log(n) / Math.Log(2)))) ==
       (int)(Math.Floor(((Math.Log(n) / Math.Log(2)))));
  }

  // Recursive function to find
  // steps needed to reduce
  // a given integer to 1
  static int reduceToOne(int N)
  {

      // Base Condition
      if(N == 1){
          return 0;
      }

      // If the number is a power of 2
      if(isPowerOfTwo(N) == true){
          return 1 + reduceToOne(N/2);
      }

      // Else subtract the greatest
      //power of 2 smaller than N
      //from N
      else{
          return 1 + reduceToOne(N - highestPowerof2(N));
      }
  }

  // Driver Code
  public static void Main()
  {

      // Input
      int N = 7;

      // Function call
      Console.Write(reduceToOne(N));
  }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
// Javascript program for the above approach

// Utility function to check
// if n is power of 2
function isPowerOfTwo(n)
{
    if (n == 0)
        return false;

    return parseInt( (Math.ceil((Math.log(n) / Math.log(2))))) == parseInt( (Math.floor(((Math.log(n) / Math.log(2))))));
}

// Utility function to find highest
// power of 2 less than or equal
// to given number
function highestPowerof2(n)
{
    let p = parseInt(Math.log(n) / Math.log(2), 10);
    return Math.pow(2, p);
}

// Recursive function to find
// steps needed to reduce
// a given integer to 1
function reduceToOne(N)
{
    // Base Condition
    if(N == 1){
        return 0;
    }
    // If the number is a power of 2
    if(isPowerOfTwo(N) == true){
        return 1 + reduceToOne(N/2);
    }
    // Else subtract the greatest
    //power of 2 smaller than N
    //from N
    else{
        return 1 + reduceToOne(N - highestPowerof2(N));
    }
}

// Driver Code

// Input
let N = 7;

// Function call
document.write(reduceToOne(N));

// This code is contributed by Samim Hossain Mondal
</script>

Output

2

Method 2 – 

Approach: The given problem can be solved using bitwise operators. Follow the steps below to solve the problem:

• Initialize two variables say res with value 0 and MSB with value log₂(N) to store the number of steps needed to reduce the number to 1 and the most significant bit of N.
• Iterate while N is not equal to 1:
  • Check if the number is the power of 2 then divide N by 2.
  • Otherwise, subtract the largest power of 2 less than N from N i.e update N as N=N-2^MSB.
  • Increment res by 1 and decrement MSB by 1.
• Finally, print res.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find steps needed to
// reduce a given integer to 1
int reduceToOne(int N)
{
    // Stores the most
    // significant bit of N
    int MSB = log2(N);

    // Stores the number of steps
    // required to reduce N to 1
    int res = 0;

    // Iterates while N
    // is not equal 1
    while (N != 1) {

        // Increment res by 1
        res++;

        // If N is power of 2
        if (N & (N - 1) == 0) {

            // Divide N by 2
            N /= 2;
        }
        // Otherwise
        else {

            // Subtract 2 ^ MSB
            // from N
            N -= (1 << MSB);
        }
        // Decrement MSB by 1
        MSB--;
    }
    // Returns res
    return res;
}

// Driver code
int main()
{
    // Input
    int N = 7;
    // Function call
    cout << reduceToOne(N) << endl;
}

Java

/*package whatever //do not write package name here */
import java.io.*;

class GFG {
    public static int logarithm(int number, int base)
    {
        int res = (int)(Math.log(number) / Math.log(base));
        return res;
    }
    public static int reduceToOne(int N)
    {

        // Stores the most
        // significant bit of N
        int MSB = logarithm(N, 2);

        // Stores the number of steps
        // required to reduce N to 1
        int res = 0;

        // Iterates while N
        // is not equal 1
        while (N != 1) {

            // Increment res by 1
            res++;

            // If N is power of 2
            if ((N & (N - 1)) == 0) {

                // Divide N by 2
                N /= 2;
            }
            // Otherwise
            else {

                // Subtract 2 ^ MSB
                // from N
                N -= (1 << MSB);
            }
            // Decrement MSB by 1
            MSB--;
        }
        // Returns res
        return res;
    }

    public static void main(String[] args)
    {
        int N = 7;
        int res = reduceToOne(N);
        System.out.println(res);
    }
}

// This code is contributed by lokeshpotta20.

Python3

# Python program for the above approach
import math

# Function to find steps needed to
# reduce a given integer to 1
def reduceToOne(N):

    # Stores the most
    # significant bit of N
    MSB = math.floor(math.log2(N))

    # Stores the number of steps
    # required to reduce N to 1
    res = 0

    # Iterates while N
    # is not equal 1
    while (N != 1):

        # Increment res by 1
        res += 1

        # If N is power of 2
        if (N & (N - 1) == 0):

            # Divide N by 2
            N //= 2

        # Otherwise
        else:
            # Subtract 2 ^ MSB
            # from N
            N -= (1 << MSB)

        # Decrement MSB by 1
        MSB-=1

    # Returns res
    return res

# Driver code
# Input
N = 7
# Function call
print(reduceToOne(N))

# This code is contributed by shubham Singh

C#

// C# program for the above approach

using System;
using System.Collections.Generic;

class GFG{

// Function to find steps needed to
// reduce a given integer to 1
static int reduceToOne(int N)
{

    // Stores the most
    // significant bit of N
    int MSB = (int)(Math.Log(N)/Math.Log(2));

    // Stores the number of steps
    // required to reduce N to 1
    int res = 0;

    // Iterates while N
    // is not equal 1
    while (N != 1) {

        // Increment res by 1
        res++;

        // If N is power of 2
        if ((N & (N - 1)) == 0) {

            // Divide N by 2
            N /= 2;
        }
        // Otherwise
        else {

            // Subtract 2 ^ MSB
            // from N
            N -= (1 << MSB);
        }

        // Decrement MSB by 1
        MSB--;
    }

    // Returns res
    return res;
}

// Driver code
public static void Main()
{
    // Input
    int N = 7;

    // Function call
    Console.Write(reduceToOne(N));
}
}

// This code is contributed by bgangwar59.

Javascript

<script>

// JavaScript program for the above approach

    function logarithm(number, base)
    {
        let res = (Math.log(number) / Math.log(base));
        return res;
    }
    function reduceToOne(N)
    {

        // Stores the most
        // significant bit of N
        let MSB = logarithm(N, 2);

        // Stores the number of steps
        // required to reduce N to 1
        let res = 0;

        // Iterates while N
        // is not equal 1
        while (N != 1) {

            // Increment res by 1
            res++;

            // If N is power of 2
            if ((N & (N - 1)) == 0) {

                // Divide N by 2
                N /= 2;
            }
            // Otherwise
            else {

                // Subtract 2 ^ MSB
                // from N
                N -= (1 << MSB);
            }
            // Decrement MSB by 1
            MSB--;
        }
        // Returns res
        return res;
    }

        // Driver Code

        let N = 7;
        let res = reduceToOne(N);
        document.write(res);

</script>

Output

2

Time Complexity: O(log(N))
Auxiliary Space: O(1)