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

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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 <bits/stdc++.h>
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[0]);
 
    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

<?php
// PHP implementation of the above approach
 
// Function to find the nearest power of 2
function nextPowerOf2($n)
{
 
    // The number
    $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
function memoryUsed(&$arr, $n)
{
    // Sum of array
    $sum = 0;
 
    // Traverse and find the sum of array
    for ($i = 0; $i < $n; $i++)
        $sum += $arr[$i];
 
    // Function call to find the 
    // nearest power of 2
    $nearest = nextPowerOf2($sum);
 
    return $nearest;
}
 
// Driver Code
$arr = array(1, 2, 3, 2);
$n = sizeof($arr);
 
echo(memoryUsed($arr, $n));
 
// This code is contributed 
// by Shivi_Aggarwal
?>

Javascript

<script>
 
// Javascript implementation of the above approach
 
// Function to find the nearest power of 2
function nextPowerOf2(n)
{
 
    // The number
    let 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
function memoryUsed(arr, n)
{
    // Sum of array
    let sum = 0;
 
    // Traverse and find the sum of array
    for (let i = 0; i < n; i++)
        sum += arr[i];
 
    // Function call to find the nearest power of 2
    let nearest = nextPowerOf2(sum);
 
    return nearest;
}
// Driver Code
let arr = [ 1, 2, 3, 2 ];
let n = arr.length;
 
document.write(memoryUsed(arr, n));
 
</script>

Output:

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.

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