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

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  • Last Updated : 12 Nov, 2021
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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:
 

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)


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