Count total unset bits in all the numbers from 1 to N

Given a positive integer N, the task is to count the total number of unset bits in the binary representation of all the numbers from 1 to N. Note that leading zeroes will not be counted as unset bits.
Examples:

Input: N = 5 
Output: 4 

Integer	Binary Representation	Count of unset bits
1	1	0
2	10	1
3	11	0
4	100	2
5	101	1

0 + 1 + 0 + 2 + 1 = 4
Input: N = 14 
Output: 17 

Approach:

1. Iterate the loop from 1 to N.
2. While number is greater than 0 divide it by 2 and check the remainder.
3. If remainder is equal to 0 then increase the value of count by 1.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the count of unset
// bits in the binary representation of
// all the numbers from 1 to n
int countUnsetBits(int n)
{

    // To store the count of unset bits
    int cnt = 0;

    // For every integer from the range [1, n]
    for (int i = 1; i <= n; i++) {

        // A copy of the current integer
        int temp = i;

        // Count of unset bits in
        // the current integer
        while (temp) {

            // If current bit is unset
            if (temp % 2 == 0)
                cnt++;

            temp = temp / 2;
        }
    }
    return cnt;
}

// Driver code
int main()
{
    int n = 5;

    cout << countUnsetBits(n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

    // Function to return the count of unset
    // bits in the binary representation of
    // all the numbers from 1 to n
    static int countUnsetBits(int n)
    {

        // To store the count of unset bits
        int cnt = 0;

        // For every integer from the range [1, n]
        for (int i = 1; i <= n; i++)
        {

            // A copy of the current integer
            int temp = i;

            // Count of unset bits in
            // the current integer
            while (temp > 0)
            {

                // If current bit is unset
                if (temp % 2 == 0)
                {
                    cnt = cnt + 1;
                }

                temp = temp / 2;
            }
        }
        return cnt;
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 5;
        System.out.println(countUnsetBits(n));
    }
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach

# Function to return the count of unset
# bits in the binary representation of
# all the numbers from 1 to n
def countUnsetBits(n) :

    # To store the count of unset bits
    cnt = 0;

    # For every integer from the range [1, n]
    for i in range(1, n + 1) :

        # A copy of the current integer
        temp = i;

        # Count of unset bits in
        # the current integer
        while (temp) :

            # If current bit is unset
            if (temp % 2 == 0) :
                cnt += 1;

            temp = temp // 2;

    return cnt;

# Driver code
if __name__ == "__main__" :

    n = 5;

    print(countUnsetBits(n));

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to return the count of unset
    // bits in the binary representation of
    // all the numbers from 1 to n
    static int countUnsetBits(int n)
    {

        // To store the count of unset bits
        int cnt = 0;

        // For every integer from the range [1, n]
        for (int i = 1; i <= n; i++)
        {

            // A copy of the current integer
            int temp = i;

            // Count of unset bits in
            // the current integer
            while (temp > 0)
            {

                // If current bit is unset
                if (temp % 2 == 0)
                    cnt = cnt + 1;

                temp = temp / 2;
            }
        }
        return cnt;
    }

    // Driver code
    public static void Main()
    {
        int n = 5;
        Console.Write(countUnsetBits(n));
    }
}

// This code is contributed by Sanjit_Prasad

Javascript

<script>

// Javascript implementation of the approach

// Function to return the count of unset
// bits in the binary representation of
// all the numbers from 1 to n
function countUnsetBits(n)
{

    // To store the count of unset bits
    var cnt = 0;

    // For every integer from the range [1, n]
    for (var i = 1; i <= n; i++) {

        // A copy of the current integer
        var temp = i;

        // Count of unset bits in
        // the current integer
        while (temp) {

            // If current bit is unset
            if (temp % 2 == 0)
                cnt++;

            temp = parseInt(temp / 2);
        }
    }
    return cnt;
}

// Driver code
var n = 5;
document.write( countUnsetBits(n));

</script>

Output:

4

Time Complexity: O(n * log n)

Auxiliary Space: O(1)