Count of total bits toggled/flipped in binary representation of 0 to N

Given an integer N, the task is to find the total number of bits toggled to obtain all numbers from 0 to N sequentially.

Examples:

Input: N = 5
Output: 8
Explanation:
Let’s represent numbers from 0 to 5 in binary:
000 -> 001 : 1 bit toggled
001 -> 010 : 2 bits toggled
010 -> 011 : 1 bit toggled
011 -> 100 : 3 bits toggled
100 -> 101 : 1 bit toggled
Hence, 8 bits toggled

Input: N = 1
Output: 1

Approach:
Follow the steps below to solve the problems:



  1. The following observations need to be made to solve the problem:

    The rightmost bit toggles every time.
    (000) -> (001) -> (010) -> (011)
    So, the contribution of this bit to the count will be N.
    The next bit toggles after every 2 numbers.
    (000) -> (001) -> (010) -> (011) -> (100)
    Hence, the contribution of this bit to the count will be N/2.

  2. Thus, we can conclude that the i th least significant bit will contribute N/(2 i) to the count of toggles.
  3. Hence, the sum of N/(2i) where i is in the range [0, log2N], gives the required answer.
  4. Hence, initialize a variable ans. Add N to ans and update N to N/2. Repeat this process until N becomes 0, to get the final result.
  5. Below is the implementation of the above approach:

    C++

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // C++ program to count
    // the number of toggles
    // required to generate
    // all numbers from 0 to N
      
    #include <bits/stdc++.h>
    using namespace std;
      
    typedef long long int ll;
      
    // Function to count and print
    // the required number of
    // toggles
    void solve(ll N)
    {
        // Store the count
        // of toggles
        ll ans = 0;
      
        while (N != 0) {
      
            // Add the contribution
            // of the current LSB
            ans += N;
      
            // Update N
            N /= 2;
        }
        // Print the result
        cout << ans << endl;
    }
    // Driver code
    int main()
    {
        ll N = 5;
        solve(N);
        return 0;
    }

    chevron_right

    
    

    Java

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // Java program to count the 
    // number of toggles required
    // to generate all numbers 
    // from 0 to N
    class GFG{
      
    // Function to count and print
    // the required number of
    // toggles
    static void solve(int N)
    {
          
        // Store the count
        // of toggles
        int ans = 0;
      
        while (N != 0)
        {
              
            // Add the contribution
            // of the current LSB
            ans += N;
      
            // Update N
            N /= 2;
        }
          
        // Print the result
        System.out.println(ans);
    }
      
    // Driver code
    public static void main(String []args)
    {
        int N = 5;
          
        solve(N);
    }
    }
      
    // This code is contributed by Ritik Bansal

    chevron_right

    
    

    Output:

    8
    

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

    Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




    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 : btc_148