# Smallest power of 2 which is greater than or equal to sum of array elements

• Difficulty Level : Basic
• Last Updated : 14 Jun, 2022

Given an array of N numbers where values of the array represent memory sizes. The memory that is required by the system can only be represented in powers of 2. The task is to return the size of the memory required by the system.
Examples:

```Input: a[] = {2, 1, 4, 5}
Output: 16
The sum of memory required is 12,
hence the nearest power of 2 is 16.

Input: a[] = {1, 2, 3, 2}
Output: 8```

Source: Microsoft Interview

Approach: The problem is a combination of summation of array elements and smallest power of 2 greater than or equal to N. Find the sum of array elements and then find the smallest power of 2 greater than or equal to N.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach``#include ``using` `namespace` `std;` `// Function to find the nearest power of 2``int` `nextPowerOf2(``int` `n)``{` `    ``// The number``    ``int` `p = 1;` `    ``// If already a power of 2``    ``if` `(n && !(n & (n - 1)))``        ``return` `n;` `    ``// Find the next power of 2``    ``while` `(p < n)``        ``p <<= 1;` `    ``return` `p;``}` `// Function to find the memory used``int` `memoryUsed(``int` `arr[], ``int` `n)``{``    ``// Sum of array``    ``int` `sum = 0;` `    ``// Traverse and find the sum of array``    ``for` `(``int` `i = 0; i < n; i++)``        ``sum += arr[i];` `    ``// Function call to find the nearest power of 2``    ``int` `nearest = nextPowerOf2(sum);` `    ``return` `nearest;``}``// Driver Code``int` `main()``{``    ``int` `arr[] = { 1, 2, 3, 2 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << memoryUsed(arr, n);` `    ``// getchar();``    ``return` `0;``}`

## Java

 `// Java implementation of the above approach` `class` `GFG``{``    ``// Function to find the nearest power of 2``    ``static` `int` `nextPowerOf2(``int` `n)``    ``{``    ` `        ``// The number``        ``int` `p = ``1``;``    ` `        ``// If already a power of 2``        ``if``(n!=``0` `&& ((n&(n-``1``)) == ``0``))``            ``return` `n;``    ` `        ``// Find the next power of 2``        ``while` `(p < n)``            ``p <<= ``1``;``    ` `        ``return` `p;``    ``}``    ` `    ``// Function to find the memory used``    ``static` `int` `memoryUsed(``int` `arr[], ``int` `n)``    ``{``        ``// Sum of array``        ``int` `sum = ``0``;``    ` `        ``// Traverse and find the sum of array``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``sum += arr[i];``    ` `        ``// Function call to find the nearest power of 2``        ``int` `nearest = nextPowerOf2(sum);``    ` `        ``return` `nearest;``    ``}``    ``// Driver Code``    ``public` `static` `void` `main(String []args)``    ``{``        ``int` `arr[] = { ``1``, ``2``, ``3``, ``2` `};``        ``int` `n = arr.length;``    ` `        ``System.out.println(memoryUsed(arr, n));``    `  `    ``}``}` `// This code is contributed``// by ihritik`

## Python3

 `# Python3 implementation of the above approach` `# Function to find the nearest power of 2``def` `nextPowerOf2(n):``    ` `    ``# The number``    ``p ``=` `1``    ` `    ``# If already a power of 2``    ``if` `(n ``and` `not``(n & (n ``-` `1``))):``        ``return` `n``        ` `    ``# Find the next power of 2``    ``while` `(p < n):``        ``p <<``=` `1``    ``return` `p` `# Function to find the memory used``def` `memoryUsed(arr, n):``    ` `    ``# Sum of array``    ``sum` `=` `0` `    ``# Traverse and find the sum of array``    ``for` `i ``in` `range``(n):``        ``sum` `+``=` `arr[i]` `    ``# Function call to find the nearest``    ``# power of 2``    ``nearest ``=` `nextPowerOf2(``sum``)` `    ``return` `nearest` `# Driver Code``arr ``=` `[``1``, ``2``, ``3``, ``2``]``n ``=` `len``(arr)``print``(memoryUsed(arr, n))` `# This code is contributed by sahishelangia`

## C#

 `// C# implementation of the above approach` `using` `System;``class` `GFG``{``    ``// Function to find the nearest power of 2``    ``static` `int` `nextPowerOf2(``int` `n)``    ``{``    ` `        ``// The number``        ``int` `p = 1;``    ` `        ``// If already a power of 2``        ``if``(n!=0 && ((n&(n-1)) == 0))``            ``return` `n;``    ` `        ``// Find the next power of 2``        ``while` `(p < n)``            ``p <<= 1;``    ` `        ``return` `p;``    ``}``    ` `    ``// Function to find the memory used``    ``static` `int` `memoryUsed(``int` `[]arr, ``int` `n)``    ``{``        ``// Sum of array``        ``int` `sum = 0;``    ` `        ``// Traverse and find the sum of array``        ``for` `(``int` `i = 0; i < n; i++)``            ``sum += arr[i];``    ` `        ``// Function call to find the nearest power of 2``        ``int` `nearest = nextPowerOf2(sum);``    ` `        ``return` `nearest;``    ``}``    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]arr = { 1, 2, 3, 2 };``        ``int` `n = arr.Length;``    ` `        ``Console.WriteLine(memoryUsed(arr, n));``    `  `    ``}``}` `// This code is contributed``// by ihritik`

## PHP

 ``

## Javascript

 ``

Output:

`8`

Time Complexity: O(N), as we are using a loop to traverse N times, where N is the number of elements in the array.

Auxiliary Space: O(1), as we are not using any extra space.

My Personal Notes arrow_drop_up