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Sum of numbers with exactly 2 bits set

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  • Difficulty Level : Medium
  • Last Updated : 31 May, 2022

Given a number n. Find the sum of all numbers up to n whose 2 bits are set.
 

Examples: 

Input : 10
Output : 33
3 + 5 + 6 + 9 + 10 = 33

Input : 100
Output : 762

Naive Approach: Find each number up to n whose 2 bits are set. If its 2 bits are set add it to the sum.
 

C++




// CPP program to find sum of numbers
// upto n whose 2 bits are set
#include <bits/stdc++.h>
using namespace std;
 
// To count number of set bits
int countSetBits(int n)
{
    int count = 0;
    while (n) {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
// To calculate sum of numbers
int findSum(int n)
{
    int sum = 0;
 
    // To count sum of number
    // whose 2 bit are set
    for (int i = 1; i <= n; i++)
        if (countSetBits(i) == 2)
            sum += i;
 
    return sum;
}
 
// Driver program to test above function
int main()
{
    int n = 10;
    cout << findSum(n);
    return 0;
}

Java



// Java program to find sum of numbers
// upto n whose 2 bits are set
public class Main {
 
    // To count number of set bits
    static int countSetBits(int n)
    {
        int count = 0;
        while (n > 0) {
            n &= (n - 1);
            count++;
        }
        return count;
    }
 
    // To calculate sum of numbers
    static int findSum(int n)
    {
        int sum = 0;
 
        // To count sum of number
        // whose 2 bit are set
        for (int i = 1; i <= n; i++)
            if (countSetBits(i) == 2)
                sum += i;
 
        return sum;
    }
 
    // Driver program to test above function
    public static void main(String[] args)
    {
        int n = 10;
 
        System.out.println(findSum(n));
    }
}
Python3



# Python program to find
# sum of numbers
# upto n whose 2 bits are set
 
# To count number of set bits
def countSetBits(n):
 
    count = 0
    while (n):
        n =n & (n - 1)
        count=count + 1
     
    return count
 
# To calculate sum of numbers
def findSum(n):
 
    sum = 0
  
    # To count sum of number
    # whose 2 bit are set
    for i in range(1,n+1):
        if (countSetBits(i) == 2):
            sum =sum + i
  
    return sum
 
# Driver code
n = 10
print(findSum(n))
 
# This code is contributed
# by Anant Agarwal.
C#



// C# program to find sum of
// numbers upto n whose 2
// bits are set
using System;
 
class GFG
{
     
    // To count number
    // of set bits
    static int countSetBits(int n)
    {
        int count = 0;
        while (n > 0)
        {
            n = n & (n - 1);
            count++;
        }
        return count;
    }
 
    // To calculate
    // sum of numbers
    static int findSum(int n)
    {
        int sum = 0;
 
        // To count sum of number
        // whose 2 bit are set
        for (int i = 1; i <= n; i++)
            if (countSetBits(i) == 2)
                sum += i;
 
        return sum;
    }
 
    // Driver Code
    static public void Main ()
    {
        int n = 10;
 
        Console.WriteLine(findSum(n));
    }
}
 
// This code is contributed by aj_36
PHP



<?php
// PHP program to find sum of numbers
// upto n whose 2 bits are set
 
// To count number of set bits
function countSetBits($n)
{
    $count = 0;
    while ($n)
    {
        $n &= ($n - 1);
        $count++;
    }
    return $count;
}
 
// To calculate sum of numbers
function findSum($n)
{
    $sum = 0;
 
    // To count sum of number
    // whose 2 bit are set
    for ($i = 1; $i <= $n; $i++)
        if (countSetBits($i) == 2)
            $sum += $i;
 
    return $sum;
}
 
    // Driver Code
    $n = 10;
    echo findSum($n);
     
// This code is contributed by anuj_67.
?>
Javascript



<script>
 
// Javascript program to find sum of numbers
// upto n whose 2 bits are set
 
// To count number of set bits
function countSetBits(n)
{
    let count = 0;
    while (n) {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
// To calculate sum of numbers
function findSum(n)
{
    let sum = 0;
 
    // To count sum of number
    // whose 2 bit are set
    for (let i = 1; i <= n; i++)
        if (countSetBits(i) == 2)
            sum += i;
 
    return sum;
}
 
// Driver program to test above function
 
    let n = 10;
    document.write(findSum(n));
     
 
// This code is contributed by Mayank Tyagi
</script>
Output: 
33
Time Complexity : O(n)
Space Complexity : O(1)
Efficient Approach: The number whose 2 bits are set is of the form 2^x + 2^y and this number is less than n. So we have to find only numbers in the range up to n which is of form 2^i + 2^j where i > 0 and 2^i < n and 0 <= j < i.
 
C++



// C++ program to find sum of numbers
// upto n whose 2 bits are set
#include <bits/stdc++.h>
using namespace std;
 
// To calculate sum of numbers
int findSum(int n)
{
    int sum = 0;
 
    // Find numbers whose 2 bits are set
    for (int i = 1; (1 << i) < n; i++) {
        for (int j = 0; j < i; j++) {
            int num = (1 << i) + (1 << j);
 
            // If number is greater then n
            // we don't include this in sum
            if (num <= n)
                sum += num;
        }
    }
 
    // Return sum of numbers
    return sum;
}
 
// Driver program to test findSum()
int main()
{
    int n = 10;
    cout << findSum(n);
    return 0;
}
Java



// Java program to find sum of numbers
// upto n whose 2 bits are set
public class Main {
 
    // To calculate sum of numbers
    static int findSum(int n)
    {
        int sum = 0;
 
        // Find numbers whose 2 bits are set
        for (int i = 1; 1 << i < n; i++) {
            for (int j = 0; j < i; j++) {
                int num = (1 << i) + (1 << j);
 
                // If number is greater then n
                // we don't include this in sum
                if (num <= n)
                    sum += num;
            }
        }
 
        // Return sum of numbers
        return sum;
    }
 
    // Driver program to test findSum()
    public static void main(String[] args)
    {
        int n = 10;
        System.out.println(findSum(n));
    }
}
Python3



# Python3 program to find sum of
# numbers upto n whose 2 bits are set
 
# To calculate sum of numbers
def findSum(n) :
 
    sum = 0
 
    # Find numbers whose 2
    # bits are set
    i = 1
    while((1 << i) < n ) :
        for j in range(0, i) :
            num = (1 << i) + (1 << j)
 
            # If number is greater then n
            # we don't include this in sum
            if (num <= n) :
                sum += num
         
        i += 1
         
    # Return sum of numbers
    return sum
 
# Driver Code
n = 10
print(findSum(n))
 
# This code is contributed
# by Smitha
C#



// C# program to find sum of numbers
// upto n whose 2 bits are set
using System;
 
public class main {
 
    // To calculate sum of numbers
    static int findSum(int n)
    {
        int sum = 0;
 
        // Find numbers whose 2 bits are set
        for (int i = 1; 1 << i < n; i++)
        {
            for (int j = 0; j < i; j++)
            {
                int num = (1 << i) + (1 << j);
                           
                // If number is greater then n
                // we don't include this in sum
                if (num <= n)
                    sum += num;
            }
        }
 
        // Return sum of numbers
        return sum;
    }
 
    // Driver Code
    public static void Main(String []args)
    {
        int n = 10;
        Console.WriteLine(findSum(n));
    }
}
 
// This Code is contributed by vt_m.
PHP



<?php
<?php
// PHP program to find sum of numbers
// upto n whose 2 bits are set
 
// To calculate sum of numbers
function findSum($n)
{
    $sum = 0;
 
    // Find numbers whose 2 bits are set
    for ($i = 1; (1 << $i) < $n; $i++)
    {
        for ($j = 0; $j < $i; $j++)
        {
            $num = (1 << $i) + (1 << $j);
 
            // If number is greater then n
            // we don't include this in sum
            if ($num <= $n)
                $sum += $num;
        }
    }
 
    // Return sum of numbers
    return $sum;
}
 
// Driver Code
$n = 10;
echo findSum($n);
 
// This code is contributed by Ajit
?>
Javascript



<script>
    // Javascript program to find sum of numbers
    // upto n whose 2 bits are set
     
    // To calculate sum of numbers
    function findSum(n)
    {
        let sum = 0;
  
        // Find numbers whose 2 bits are set
        for (let i = 1; 1 << i < n; i++)
        {
            for (let j = 0; j < i; j++)
            {
                let num = (1 << i) + (1 << j);
                            
                // If number is greater then n
                // we don't include this in sum
                if (num <= n)
                    sum += num;
            }
        }
  
        // Return sum of numbers
        return sum;
    }
     
    let n = 10;
      document.write(findSum(n));
     
</script>
Output :  
33
Time Complexity : O((log n)*(log n))
Space Complexity : O(1)
This article is contributed by nuclode. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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