# 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 = log2
And the sum of all numbers which are perfect power of 2 is given by 2r – 1

1.
2. 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++

 `// C++ implementation of the` `// approach` `#include ` `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;` `}`

## Java

 `// 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`

## Python3

 `# 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`

## C#

 `// 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`

## Javascript

 ``

Output:

`0`

Time Complexity: O(1)

Auxiliary Space: O(1)

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