# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program to count ` `// the number of toggles ` `// required to generate ` `// all numbers from 0 to N ` ` `  `#include ` `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; ` `} `

## Java

 `// 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 `

Output:

```8
```

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

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