Convert given integer X to the form 2^N – 1

Given an integer x. The task is to convert x to the form 2n – 1 by performing the following operations in specified order on x:

  1. You can select any non-negative integer n and update x = x xor (2n – 1)
  2. Replace x with x + 1.

The first applied operation must be of first type, the second of second type, the third again of first type and so on. Formally, if we number the operations from one in the order they are executed, then odd-numbered operations must be of the first type and the even-numbered operations must be of second type. The task is to find the number of operation required to convert x to the form 2n – 1.

Examples:



Input: x = 39
Output: 4
Operation 1: Pick n = 5, x is transformed into (39 xor 31) = 56.
Operation 2: x = 56 + 1 = 57
Operation 3: Pick n = 3, x is transformed into (57 xor 7) = 62.
Operation 4: x = 62 + 1 = 63 i.e. (26 – 1).
So, total number of operations are 4.

Input: x = 7
Output: 0
As 23 -1 = 7.
So, no operation is required.

Approach: Take the smallest number larger than x which is of the form of 2n – 1 say num, then update x = x xor num and then x = x + 1 performing two operations. Repeat the step until x is of the form 2n – 1. Print the number of operations performed in the end.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
const int MAX = 24;
  
// Function to return the count
// of operations required
int countOp(int x)
{
  
    // To store the powers of 2
    int arr[MAX];
    arr[0] = 1;
    for (int i = 1; i < MAX; i++)
        arr[i] = arr[i - 1] * 2;
  
    // Temporary variable to store x
    int temp = x;
  
    bool flag = true;
  
    // To store the index of
    // smaller number larger than x
    int ans;
  
    // To store the count of operations
    int operations = 0;
  
    bool flag2 = false;
  
    for (int i = 0; i < MAX; i++) {
  
        if (arr[i] - 1 == x)
            flag2 = true;
  
        // Stores the index of number
        // in the form of 2^n - 1
        if (arr[i] > x) {
            ans = i;
            break;
        }
    }
  
    // If x is already in the form
    // 2^n - 1 then no operation is required
    if (flag2)
        return 0;
  
    while (flag) {
  
        // If number is less than x increase the index
        if (arr[ans] < x)
            ans++;
  
        operations++;
  
        // Calculate all the values (x xor 2^n-1)
        // for all possible n
        for (int i = 0; i < MAX; i++) {
            int take = x ^ (arr[i] - 1);
            if (take <= arr[ans] - 1) {
  
                // Only take value which is
                // closer to the number
                if (take > temp)
                    temp = take;
            }
        }
  
        // If number is in the form of 2^n - 1 then break
        if (temp == arr[ans] - 1) {
            flag = false;
            break;
        }
  
        temp++;
        operations++;
        x = temp;
  
        if (x == arr[ans] - 1)
            flag = false;
    }
  
    // Return the count of operations
    // required to obtain the number
    return operations;
}
  
// Driver code
int main()
{
    int x = 39;
  
    cout << countOp(x);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
import java.io.*;
  
class GFG
{
  
static int MAX = 24;
  
// Function to return the count
// of operations required
static int countOp(int x)
{
  
    // To store the powers of 2
    int arr[] = new int[MAX];
    arr[0] = 1;
    for (int i = 1; i < MAX; i++)
        arr[i] = arr[i - 1] * 2;
  
    // Temporary variable to store x
    int temp = x;
  
    boolean flag = true;
  
    // To store the index of
    // smaller number larger than x
    int ans =0;
  
    // To store the count of operations
    int operations = 0;
  
    boolean flag2 = false;
  
    for (int i = 0; i < MAX; i++) 
    {
  
        if (arr[i] - 1 == x)
            flag2 = true;
  
        // Stores the index of number
        // in the form of 2^n - 1
        if (arr[i] > x) 
        {
            ans = i;
            break;
        }
    }
  
    // If x is already in the form
    // 2^n - 1 then no operation is required
    if (flag2)
        return 0;
  
    while (flag) 
    {
  
        // If number is less than x increase the index
        if (arr[ans] < x)
            ans++;
  
        operations++;
  
        // Calculate all the values (x xor 2^n-1)
        // for all possible n
        for (int i = 0; i < MAX; i++) 
        {
            int take = x ^ (arr[i] - 1);
            if (take <= arr[ans] - 1
            {
  
                // Only take value which is
                // closer to the number
                if (take > temp)
                    temp = take;
            }
        }
  
        // If number is in the form of 2^n - 1 then break
        if (temp == arr[ans] - 1
        {
            flag = false;
            break;
        }
  
        temp++;
        operations++;
        x = temp;
  
        if (x == arr[ans] - 1)
            flag = false;
    }
  
    // Return the count of operations
    // required to obtain the number
    return operations;
}
  
    // Driver code
    public static void main (String[] args) 
    {
        int x = 39;
        System.out.println(countOp(x));
    }
}
  
// This code is contributed by anuj_67..

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
  
MAX = 24
  
# Function to return the count 
# of operations required 
def countOp(x) : 
  
    # To store the powers of 2 
    arr = [0]*MAX
    arr[0] = 1
    for i in range(1, MAX) : 
        arr[i] = arr[i - 1] * 2
  
    # Temporary variable to store x 
    temp = x; 
  
    flag = True
  
    # To store the index of 
    # smaller number larger than x 
    ans = 0
  
    # To store the count of operations 
    operations = 0
  
    flag2 = False
  
    for i in range(MAX) :
  
        if (arr[i] - 1 == x) :
            flag2 = True
  
        # Stores the index of number 
        # in the form of 2^n - 1 
        if (arr[i] > x) :
            ans = i; 
            break
      
    # If x is already in the form 
    # 2^n - 1 then no operation is required 
    if (flag2) :
        return 0
  
    while (flag) : 
  
        # If number is less than x increase the index 
        if (arr[ans] < x) :
            ans += 1
  
        operations += 1
  
        # Calculate all the values (x xor 2^n-1) 
        # for all possible n 
        for i in range(MAX) :
            take = x ^ (arr[i] - 1); 
              
            if (take <= arr[ans] - 1) :
  
                # Only take value which is 
                # closer to the number 
                if (take > temp) :
                    temp = take; 
  
        # If number is in the form of 2^n - 1 then break 
        if (temp == arr[ans] - 1) : 
            flag = False
            break
  
        temp += 1
        operations += 1
        x = temp; 
  
        if (x == arr[ans] - 1) :
            flag = False
  
    # Return the count of operations 
    # required to obtain the number 
    return operations; 
  
  
# Driver code 
if __name__ == "__main__"
  
    x = 39
  
    print(countOp(x)); 
  
    # This code is contributed by AnkitRai01

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
  
class GFG
{
      
static int MAX = 24;
  
// Function to return the count
// of operations required
static int countOp(int x)
{
  
    // To store the powers of 2
    int []arr = new int[MAX];
    arr[0] = 1;
    for (int i = 1; i < MAX; i++)
        arr[i] = arr[i - 1] * 2;
  
    // Temporary variable to store x
    int temp = x;
  
    bool flag = true;
  
    // To store the index of
    // smaller number larger than x
    int ans = 0;
  
    // To store the count of operations
    int operations = 0;
  
    bool flag2 = false;
  
    for (int i = 0; i < MAX; i++) 
    {
  
        if (arr[i] - 1 == x)
            flag2 = true;
  
        // Stores the index of number
        // in the form of 2^n - 1
        if (arr[i] > x) 
        {
            ans = i;
            break;
        }
    }
  
    // If x is already in the form
    // 2^n - 1 then no operation is required
    if (flag2)
        return 0;
  
    while (flag) 
    {
  
        // If number is less than x increase the index
        if (arr[ans] < x)
            ans++;
  
        operations++;
  
        // Calculate all the values (x xor 2^n-1)
        // for all possible n
        for (int i = 0; i < MAX; i++) 
        {
            int take = x ^ (arr[i] - 1);
            if (take <= arr[ans] - 1) 
            {
  
                // Only take value which is
                // closer to the number
                if (take > temp)
                    temp = take;
            }
        }
  
        // If number is in the form of 2^n - 1 then break
        if (temp == arr[ans] - 1) 
        {
            flag = false;
            break;
        }
  
        temp++;
        operations++;
        x = temp;
  
        if (x == arr[ans] - 1)
            flag = false;
    }
  
    // Return the count of operations
    // required to obtain the number
    return operations;
}
  
// Driver code
static public void Main ()
{
      
    int x = 39;
    Console.WriteLine(countOp(x));
}
}
  
// This code is contributed by ajit.

chevron_right


Output:

4


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : vt_m, jit_t, AnkitRai01