Sum of numbers with exactly 2 bits set

Difficulty Level : Medium
Last Updated : 29 Oct, 2018

Given a number n. Find sum of all number upto n whose 2 bits are set.

Examples:

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

Input : 100
Output : 762

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

Naive Approach: Find each number upto 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. 
?> 

Output:

Efficient Approach: The number whose 2 bits are set is of the form 2^x + 2^y and this number is less then n. So we have to find only numbers in the range upto 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 
?> 

Output :

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 contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Related Articles

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

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP