Sum of first N natural numbers with all powers of 2 added twice

Given an integer N, the task is to calculate the sum of first N natural numbers adding all powers of 2 twice to the sum.
Examples: 

Input: N = 4 
Output: 17 
Explanation: 
Sum = 2+4+3+8 = 17 
Since 1, 2 and 4 are 2 0, 2 1 and 2 2 respectively, they are added twice to the sum.

Input: N = 5 
Output: 22 
Explanation: 
The sum is equal to 2+4+3+8+5 = 22, 
because 1, 2 and 4 are 2 0, 2 1 and 2 2 respectively. 
 

Naive Approach: 
The simplest approach to solve this problem is to iterate upto N, and keep calculating the sum by adding every number once except the powers of 2, which needs to be added twice. 
Time Complexity: O(N) 
Auxiliary Space: O(1)

Efficient Approach: 
Follow the steps below to optimize the above approach: 

  1. Calculate sum of first N natural numbers by the formula (N * (N + 1)) / 2.
  2. Now, all powers of 2 needs to be added once more. Sum of all powers of 2 up to N can be calculated as 2 log2(N) + 1 – 1
  3. Hence, the required sum is: 

    (N * (N + 1)) / 2 + 2 log2(N) + 1 – 1



Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to raise N to the
// power P and return the value
double power(int N, int P)
{
    return pow(N, P);
}
  
// Function to calculate the
// log base 2 of an integer
int Log2(int N)
{
  
    // Calculate log2(N) indirectly
    // using log() method
    int result = (int)(log(N) / log(2));
    return result;
}
  
// Function to calculate and
// return the required sum
double specialSum(int n)
{
  
    // Sum of first N natural
    // numbers
    double sum = n * (n + 1) / 2;
  
    // Sum of all powers of 2
    // up to N
    int a = Log2(n);
    sum = sum + power(2, a + 1) - 1;
  
    return sum;
}
  
// Driver code
int main()
{
    int n = 4;
  
    cout << (specialSum(n)) << endl;
  
    return 0;
}
  
// This code is contributed by divyeshrabadiya07

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to implement
// the above approach
import java.util.*;
import java.lang.Math;
  
class GFG {
  
    // Function to raise N to the
    // power P and return the value
    static double power(int N, int P)
    {
        return Math.pow(N, P);
    }
  
    // Function to calculate the
    // log base 2 of an integer
    public static int log2(int N)
    {
  
        // Calculate log2(N) indirectly
        // using log() method
        int result = (int)(Math.log(N)
                           / Math.log(2));
        return result;
    }
  
    // Function to calculate and
    // return the required sum
    static double specialSum(int n)
    {
  
        // Sum of first N natural
        // numbers
        double sum = n * (n + 1) / 2;
  
        // Sum of all powers of 2
        // up to N
        int a = log2(n);
        sum = sum + power(2, a + 1) - 1;
  
        return sum;
    }
  
    // Driver Code
    public static void main(String[] args)
    {
  
        int n = 4;
  
        System.out.println(specialSum(n));
    }
}

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to implement
// the above approach
using System;
class GFG
{
   
    // Function to raise N to the
    // power P and return the value
    static double power(int N, int P)
    {
        return Math.Pow(N, P);
    }
   
    // Function to calculate the
    // log base 2 of an integer
    public static int log2(int N)
    {
   
        // Calculate log2(N) indirectly
        // using log() method
        int result = (int)(Math.Log(N) /
                           Math.Log(2));
        return result;
    }
   
    // Function to calculate and
    // return the required sum
    static double specialSum(int n)
    {
   
        // Sum of first N natural
        // numbers
        double sum = (double)(n) * (n + 1) / 2;
   
        // Sum of all powers of 2
        // up to N
        int a = log2(n);
        sum = (sum) + power(2, a + 1) - 1;
   
        return sum;
    }
   
    // Driver Code
    public static void Main(string[] args)
    {
        int n = 4;
   
        Console.Write(specialSum(n));
    }
}
  
// This code is conributed by Ritik Bansal

chevron_right


Output: 

17.0

Time Complexity: O(log2(N)) 
Auxiliary Space: O(1)
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : btc_148, divyeshrabadiya07