# 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

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

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++

 `// 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 `

Output:

```0
```

Time Complexity: O(1)

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