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Sum of first N natural numbers by taking powers of 2 as negative number

Given a number N ( maybe up to 10^9 ). The task is to find the sum of first N natural number taking powers of 2 as a negative number.
Examples:

```Input: N = 4
Output: -4
- 1 - 2 + 3 - 4 = -4
1, 2, and 4 are the powers of two.

Input: N = 5
Output: 1```

Approach: An efficient solution is to store the powers of two in an array and then store presum of this array in another array. This array size can be at most 30. So, normally search for the first element in the power array which is greater than the given number.
Below is the implementation of above approach:

C++

 `// C++ implementation of above approach``#include ``using` `namespace` `std;` `// to store power of 2``int` `power[31];` `// to store presum of the power of 2's``int` `pre[31];` `// function to find power of 2``void` `PowerOfTwo()``{``    ``// to store power of 2``    ``int` `x = 1;``    ``for` `(``int` `i = 0; i < 31; i++) {``        ``power[i] = x;``        ``x *= 2;``    ``}` `    ``// to store pre sum``    ``pre[0] = 1;``    ``for` `(``int` `i = 1; i < 31; i++)``        ``pre[i] = pre[i - 1] + power[i];``}` `// Function to find the sum``int` `Sum(``int` `n)``{``    ``// first store sum of``    ``// first n natural numbers.``    ``int` `ans = n * (n + 1) / 2;` `    ``// find the first greater number than given``    ``// number then minus double of this``    ``// from answer``    ``for` `(``int` `i = 0; i < 31; i++) {``        ``if` `(power[i] > n) {``            ``ans -= 2 * pre[i - 1];``            ``break``;``        ``}``    ``}` `    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``// function call``    ``PowerOfTwo();` `    ``int` `n = 4;` `    ``// function call``    ``cout << Sum(n);` `    ``return` `0;``}`

C

 `// C implementation of above approach``#include ` `// to store power of 2``int` `power[31];` `// to store presum of the power of 2's``int` `pre[31];` `// function to find power of 2``void` `PowerOfTwo()``{``    ``// to store power of 2``    ``int` `x = 1;``    ``for` `(``int` `i = 0; i < 31; i++) {``        ``power[i] = x;``        ``x *= 2;``    ``}` `    ``// to store pre sum``    ``pre[0] = 1;``    ``for` `(``int` `i = 1; i < 31; i++)``        ``pre[i] = pre[i - 1] + power[i];``}` `// Function to find the sum``int` `Sum(``int` `n)``{``    ``// first store sum of``    ``// first n natural numbers.``    ``int` `ans = n * (n + 1) / 2;` `    ``// find the first greater number than given``    ``// number then minus double of this``    ``// from answer``    ``for` `(``int` `i = 0; i < 31; i++) {``        ``if` `(power[i] > n) {``            ``ans -= 2 * pre[i - 1];``            ``break``;``        ``}``    ``}` `    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``// function call``    ``PowerOfTwo();` `    ``int` `n = 4;` `    ``// function call``    ``printf``(``"%d"``,Sum(n));` `    ``return` `0;``}` `// This code is contributed by kothavvsaakash.`

Java

 `// Java implementation of above approach``import` `java.io.*;` `class` `GFG {``    `  `// to store power of 2``static` `int` `power[] = ``new` `int``[``31``];` `// to store presum of the power of 2's``static` `int` `pre[] = ``new` `int``[``31``];` `// function to find power of 2``static` `void` `PowerOfTwo()``{``    ``// to store power of 2``    ``int` `x = ``1``;``    ``for` `(``int` `i = ``0``; i < ``31``; i++) {``        ``power[i] = x;``        ``x *= ``2``;``    ``}` `    ``// to store pre sum``    ``pre[``0``] = ``1``;``    ``for` `(``int` `i = ``1``; i < ``31``; i++)``        ``pre[i] = pre[i - ``1``] + power[i];``}` `// Function to find the sum``static` `int` `Sum(``int` `n)``{``    ``// first store sum of``    ``// first n natural numbers.``    ``int` `ans = n * (n + ``1``) / ``2``;` `    ``// find the first greater number than given``    ``// number then minus double of this``    ``// from answer``    ``for` `(``int` `i = ``0``; i < ``31``; i++) {``        ``if` `(power[i] > n) {``            ``ans -= ``2` `* pre[i - ``1``];``            ``break``;``        ``}``    ``}` `    ``return` `ans;``}` `// Driver code``    ``public` `static` `void` `main (String[] args) {``        ` `    ``// function call``    ``PowerOfTwo();` `    ``int` `n = ``4``;` `    ``// function call``    ``System.out.println( Sum(n));``    ``}``}`` ``// This code is contributed by ajit`

Python 3

 `# Python 3 implementation of``# above approach` `# to store power of 2``power ``=` `[``0``] ``*` `31` `# to store presum of the``# power of 2's``pre ``=` `[``0``] ``*` `31` `# function to find power of 2``def` `PowerOfTwo():` `    ``# to store power of 2``    ``x ``=` `1``    ``for` `i ``in` `range``(``31``):``        ``power[i] ``=` `x``        ``x ``*``=` `2` `    ``# to store pre sum``    ``pre[``0``] ``=` `1``    ``for` `i ``in` `range``(``1``, ``31``):``        ``pre[i] ``=` `pre[i ``-` `1``] ``+` `power[i]` `# Function to find the sum``def` `Sum``(n):``    ` `    ``# first store sum of``    ``# first n natural numbers.``    ``ans ``=` `n ``*` `(n ``+` `1``) ``/``/` `2` `    ``# find the first greater number``    ``# than given number then minus``    ``# double of this from answer``    ``for` `i ``in` `range``( ``31``) :``        ``if` `(power[i] > n):``            ``ans ``-``=` `2` `*` `pre[i ``-` `1``]``            ``break` `    ``return` `ans` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``# function call``    ``PowerOfTwo()` `    ``n ``=` `4` `    ``# function call``    ``print``(``Sum``(n))` `# This code is contributed``# by ChitraNayal`

C#

 `// C# implementation of``// above approach``using` `System;``class` `GFG``{``    ` `// to store power of 2``static` `int``[] power = ``new` `int``[31];` `// to store presum of the``// power of 2's``static` `int``[] pre = ``new` `int``[31];` `// function to find power of 2``static` `void` `PowerOfTwo()``{``    ``// to store power of 2``    ``int` `x = 1;``    ``for` `(``int` `i = 0; i < 31; i++)``    ``{``        ``power[i] = x;``        ``x *= 2;``    ``}` `    ``// to store pre sum``    ``pre[0] = 1;``    ``for` `(``int` `i = 1; i < 31; i++)``        ``pre[i] = pre[i - 1] + power[i];``}` `// Function to find the sum``static` `int` `Sum(``int` `n)``{``    ``// first store sum of``    ``// first n natural numbers.``    ``int` `ans = n * (n + 1) / 2;` `    ``// find the first greater number``    ``// than given number then minus``    ``// double of this from answer``    ``for` `(``int` `i = 0; i < 31; i++)``    ``{``        ``if` `(power[i] > n)``        ``{``            ``ans -= 2 * pre[i - 1];``            ``break``;``        ``}``    ``}` `    ``return` `ans;``}` `// Driver code``public` `static` `void` `Main ()``{``    ` `    ``// function call``    ``PowerOfTwo();``    ` `    ``int` `n = 4;``    ` `    ``// function call``    ``Console.WriteLine(Sum(n));``}``}` `// This code is contributed``// by anuj_67`

PHP

 ` ``\$n``)``        ``{``            ``\$ans` `-= 2 * ``\$pre``[``\$i` `- 1];``            ``break``;``        ``}` `    ``return` `\$ans``;``}` `// Driver code` `// function call``PowerOfTwo();` `\$n` `= 4;` `// function call``print``(Sum(``\$n``));` `// This code is contributed by mits``?>`

Javascript

 ``

Output:

`-4`

Time Complexity: O(31)

Auxiliary Space: O(31)