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Count total set bits in all numbers from 1 to N | Set 3

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  • Difficulty Level : Medium
  • Last Updated : 27 Sep, 2022
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Given a positive integer N, the task is to count the total number of set bits in binary representation of all the numbers from 1 to N.

Examples: 

Input: N = 3 
Output:
setBits(1) + setBits(2) + setBits(3) = 1 + 1 + 2 = 4

Input: N = 6 
Output:

Approach: Solution to this problem has been published in the Set 1 and the Set 2 of this article. Here, a dynamic programming based approach is discussed.  

  • Base case: Number of set bits in 0 is 0.
  • For any number n: n and n>>1 has same no of set bits except for the rightmost bit.

Example: n = 11 (1011),  n >> 1 = 5 (101)… same bits in 11 and 5 are marked bold. So assuming we already know set bit count of 5, we only need to take care for the rightmost bit of 11 which is 1. setBit(11) = setBit(5) + 1 = 2 + 1 = 3

Rightmost bit is 1 for odd and 0 for even number.

Recurrence Relation: setBit(n) = setBit(n>>1) + (n & 1) and setBit(0) = 0

We can use bottom-up dynamic programming approach to solve this.

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
// set bits in all the integers
// from the range [1, n]
int countSetBits(int n)
{
 
    // To store the required count
    // of the set bits
    int cnt = 0;
 
    // To store the count of set
    // bits in every integer
    vector<int> setBits(n + 1);
 
    // 0 has no set bit
    setBits[0] = 0;
 
    for (int i = 1; i <= n; i++) {
 
        setBits[i] = setBits[i >> 1] + (i & 1);
    }
 
    // Sum all the set bits
    for (int i = 0; i <= n; i++) {
        cnt = cnt + setBits[i];
    }
 
    return cnt;
}
 
// Driver code
int main()
{
    int n = 6;
 
    cout << countSetBits(n);
 
    return 0;
}


Java




// Java implementation of the approach
import java.util.*;
 
class GFG {
 
    // Function to return the count of
    // set bits in all the integers
    // from the range [1, n]
    static int countSetBits(int n)
    {
 
        // To store the required count
        // of the set bits
        int cnt = 0;
 
        // To store the count of set
        // bits in every integer
        int[] setBits = new int[n + 1];
 
        // 0 has no set bit
        setBits[0] = 0;
 
        // For the rest of the elements
        for (int i = 1; i <= n; i++) {
 
            setBits[i] = setBits[i >> 1] + (i & 1);
        }
 
        // Sum all the set bits
        for (int i = 0; i <= n; i++) {
            cnt = cnt + setBits[i];
        }
        return cnt;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 6;
 
        System.out.println(countSetBits(n));
    }
}
 
// This code is contributed by Princi Singh


Python3




# Python3 implementation of the approach
 
# Function to return the count of
# set bits in all the integers
# from the range [1, n]
 
 
def countSetBits(n):
 
    # To store the required count
    # of the set bits
    cnt = 0
 
    # To store the count of set
    # bits in every integer
    setBits = [0 for x in range(n + 1)]
 
    # 0 has no set bit
    setBits[0] = 0
 
    # For the rest of the elements
    for i in range(1, n + 1):
        setBits[i] = setBits[i // 2] + (i & 1)
 
    # Sum all the set bits
    for i in range(0, n + 1):
        cnt = cnt + setBits[i]
 
    return cnt
 
 
# Driver code
n = 6
print(countSetBits(n))
 
# This code is contributed by Sanjit Prasad


C#




// C# implementation of the approach
using System;
 
class GFG {
 
    // Function to return the count of
    // set bits in all the integers
    // from the range [1, n]
    static int countSetBits(int n)
    {
 
        // To store the required count
        // of the set bits
        int cnt = 0;
 
        // To store the count of set
        // bits in every integer
        int[] setBits = new int[n + 1];
 
        // 0 has no set bit
        setBits[0] = 0;
 
        // 1 has a single set bit
        setBits[1] = 1;
 
        // For the rest of the elements
        for (int i = 2; i <= n; i++) {
 
            // If current element i is even then
            // it has set bits equal to the count
            // of the set bits in i / 2
            if (i % 2 == 0) {
                setBits[i] = setBits[i / 2];
            }
 
            // Else it has set bits equal to one
            // more than the previous element
            else {
                setBits[i] = setBits[i - 1] + 1;
            }
        }
 
        // Sum all the set bits
        for (int i = 0; i <= n; i++) {
            cnt = cnt + setBits[i];
        }
        return cnt;
    }
 
    // Driver code
    static public void Main()
    {
        int n = 6;
 
        Console.WriteLine(countSetBits(n));
    }
}
 
// This code is contributed by AnkitRai01


Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the count of
// set bits in all the integers
// from the range [1, n]
function countSetBits(n)
{
     
    // To store the required count
    // of the set bits
    var cnt = 0;
 
    // To store the count of set
    // bits in every integer
    var setBits = Array.from(
        {length: n + 1}, (_, i) => 0);
 
    // 0 has no set bit
    setBits[0] = 0;
 
    // 1 has a single set bit
    setBits[1] = 1;
 
    // For the rest of the elements
    for(i = 2; i <= n; i++)
    {
         
        // If current element i is even then
        // it has set bits equal to the count
        // of the set bits in i / 2
        if (i % 2 == 0)
        {
            setBits[i] = setBits[i / 2];
        }
 
        // Else it has set bits equal to one
        // more than the previous element
        else
        {
            setBits[i] = setBits[i - 1] + 1;
        }
    }
 
    // Sum all the set bits
    for(i = 0; i <= n; i++)
    {
        cnt = cnt + setBits[i];
    }
    return cnt;
}
 
// Driver code
var n = 6;
 
document.write(countSetBits(n));
 
// This code is contributed by 29AjayKumar
 
</script>


Output: 

9

 

Time Complexity: O(N), where N is the given number.
Auxiliary Space: O(N), for creating an additional array of size N + 1.

Another simple and easy-to-understand solution:

A simple easy to implement and understand solution would be not using bits operations.  The solution is to directly count set bits using __builtin_popcount(). The solution is explained in code using comments.

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count of
// set bits in all the integers
// from the range [1, n]
int countSetBits(int n)
{
 
    // To store the required count
    // of the set bits
    int cnt = 0;
 
    // Calculate set bits in each number using
    // __builtin_popcount() and  Sum all the set bits
    for (int i = 1; i <= n; i++) {
        cnt = cnt + __builtin_popcount(i);
    }
 
    return cnt;
}
 
// Driver code
int main()
{
    int n = 6;
 
    cout << countSetBits(n);
 
    return 0;
}
 
// This article is contributed by Abhishek


Java




// Java implementation of the approach
import java.util.*;
 
class GFG {
 
    // Function to return the count of
    // set bits in all the integers
    // from the range [1, n]
    static int countSetBits(int n)
    {
 
        // To store the required count
        // of the set bits
        int cnt = 0;
 
        // Calculate set bits in each number using
        // Integer.bitCount() and  Sum all the set bits
        for (int i = 1; i <= n; i++) {
            cnt = cnt + Integer.bitCount(i);
        }
 
        return cnt;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 6;
        System.out.print(countSetBits(n));
    }
}
 
// This code is contributed by Rajput-Ji


Python3




# Python implementation of the approach
 
# Function to return the count of
# set bits in all the integers
# from the range [1, n]
 
 
def countSetBits(n):
 
    # To store the required count
    # of the set bits
    cnt = 0
 
    # Calculate set bits in each number using
    # Integer.bitCount() and Sum all the set bits
    for i in range(1, n+1):
        cnt = cnt + (bin(i)[2:]).count('1')
 
    return cnt
 
 
# Driver code
if __name__ == '__main__':
    n = 6
    print(countSetBits(n))
 
# This code is contributed by Rajput-Ji


C#




// C# implementation of the approach
using System;
using System.Linq;
 
public class GFG {
 
    // Function to return the count of
    // set bits in all the integers
    // from the range [1, n]
    static int countSetBits(int n)
    {
 
        // To store the required count
        // of the set bits
        int cnt = 0;
 
        // Calculate set bits in each number using
        // int.bitCount() and  Sum all the set bits
        for (int i = 1; i <= n; i++) {
            cnt = cnt
                  + (Convert.ToString(i, 2).Count(
                      c = > c == '1'));
        }
 
        return cnt;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int n = 6;
        Console.Write(countSetBits(n));
    }
}
 
// This code is contributed by Rajput-Ji


Javascript




<script>
// javascript implementation of the approach
 
function bitCount (n) {
  n = n - ((n >> 1) & 0x55555555)
  n = (n & 0x33333333) + ((n >> 2) & 0x33333333)
  return ((n + (n >> 4) & 0xF0F0F0F) * 0x1010101) >> 24
}
    // Function to return the count of
    // set bits in all the integers
    // from the range [1, n]
    function countSetBits(n) {
 
        // To store the required count
        // of the set bits
        var cnt = 0;
 
        // Calculate set bits in each number using
        // Integer.bitCount() and Sum all the set bits
        for (i = 1; i <= n; i++) {
            cnt = cnt + bitCount(i);
        }
 
        return cnt;
    }
 
    // Driver code
        var n = 6;
        document.write(countSetBits(n));
 
// This code is contributed by Rajput-Ji
</script>


Output

9

Time Complexity: O(NlogN), where N is the given number.
Auxiliary Space: O(1)

Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.


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