Calculate sum of all integers from 1 to N, excluding perfect power of 2

Given a positive integer N, the task is to calculate the sum of all integers from 1 to N but excluding the number which is a perfect power of 2.

Examples:

Input: N = 2
Output: 0

Input: N = 1000000000
Output: 499999998352516354

Naive Approach:
The naive approach is to iterate every number from 1 to N and compute the sum in the variable by excluding the number which is a perfect power of 2. But to compute the sum to the number 10^9, the above approach will give Time Limit Error.



Time Complexity: O(N)

Efficient Approach:
To find desired sum, below are the steps:

  1. Find the sum of all the number till N using the formula discussed in this article in O(1) time.
  2. Since sum of all perfect power of 2 forms a Geometric Progression. Hence the sum of all powers of 2 less than N is calculated by the below formula:

    The number of element with perfect power of 2 less than N is given by log2N,
    Let r = log2N
    And the sum of all numbers which are perfect power of 2 is given by 2r – 1.

  3. Subtract the sum of all perfect powers of 2 calculated above from the sum of first N numbers to get the result.

Below is the implementation of the above approach:

C++

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// C++ implementation of the
// approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the required
// summation
void findSum(int N)
{
    // Find the sum of first N
    // integers using the formula
    int sum = (N) * (N + 1) / 2;
      
    int r = log2(N) + 1;
      
    // Find the sum of numbers 
    // which are exact power of
    // 2 by using the formula
    int expSum = pow(2, r) - 1;    
  
    // Print the final Sum
    cout << sum - expSum << endl;
}
  
// Driver's Code
int main()
{
    int N = 2;
  
    // Function to find the
    // sum
    findSum(N);
    return 0;
}

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Java

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// Java implementation of the above approach 
import java.lang.Math;
  
class GFG{
      
// Function to find the required 
// summation 
public static void findSum(int N) 
  
    // Find the sum of first N 
    // integers using the formula 
    int sum = (N) * (N + 1) / 2
          
    int r = (int)(Math.log(N) / 
                  Math.log(2)) + 1
          
    // Find the sum of numbers 
    // which are exact power of 
    // 2 by using the formula 
    int expSum = (int)(Math.pow(2, r)) - 1;     
      
    // Print the final Sum 
    System.out.println(sum - expSum); 
  
// Driver Code
public static void main(String[] args)
{
    int N = 2
  
    // Function to find the sum 
    findSum(N); 
}
}
  
// This code is contributed by divyeshrabadiya07

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Python3

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# Python 3 implementation of the
# approach
from math import log2,pow
  
# Function to find the required
# summation
def findSum(N):
    # Find the sum of first N
    # integers using the formula
    sum = (N) * (N + 1) // 2
      
    r = log2(N) + 1
      
    # Find the sum of numbers 
    # which are exact power of
    # 2 by using the formula
    expSum = pow(2, r) - 1
  
    # Print the final Sum
    print(int(sum - expSum))
  
# Driver's Code
if __name__ == '__main__':
    N = 2
  
    # Function to find the
    # sum
    findSum(N)
      
# This code is contributed by Surendra_Gangwar

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C#

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// C# implementation of the above approach 
using System;
  
class GFG{ 
      
// Function to find the required 
// summation 
public static void findSum(int N) 
  
    // Find the sum of first N 
    // integers using the formula 
    int sum = (N) * (N + 1) / 2; 
          
    int r = (int)(Math.Log(N) / 
                  Math.Log(2)) + 1; 
          
    // Find the sum of numbers 
    // which are exact power of 
    // 2 by using the formula 
    int expSum = (int)(Math.Pow(2, r)) - 1; 
      
    // Print the final Sum 
    Console.Write(sum - expSum); 
  
// Driver Code 
public static void Main(string[] args) 
    int N = 2; 
  
    // Function to find the sum 
    findSum(N); 
  
// This code is contributed by rutvik_56

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Output:

0

Time Complexity: O(1)

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