Count of numbers which can be made power of 2 by given operation

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Given a array arr[], the task is to count the numbers which can be made power of 2 with the following operation:
1 can be added to any element atmost once if its not already a power of 2.

Examples:

Input: arr[] = {2, 3, 7, 9, 15}
Output: 4
3, 7 and 15 can be made a power of 2 by adding 1, and 2 is already a power of 2

Input: arr[] = {5, 6, 9, 3, 1}
Output: 2

Approach: Traverse the array and check if the current element is a power of 2, if it is then update count = count + 1. If its not a power of 2 then check for one element greater i.e. arr[i] + 1. To check if an element is a power of 2:

Naive method is to repeatedly divide the element by 2 until it gives either 0 or 1 as the remainder. if the remainder is 1 then its a power of 2 else its not a power of 2.
Efficient method: If X & (X – 1) = 0 then X is a power of two.
Say, X = 16 = 10000 and X – 1 = 15 = 01111 then X & (X – 1) = 10000 & 01111 = 0 i.e. X = 16 is a power of 2.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if x is a power of 2
bool isPowerOfTwo(int x)
{
    if (x == 0)
        return false;
 
    // If x & (x-1) = 0 then x is a power of 2
    if (!(x & (x - 1)))
        return true;
    else
        return false;
}
 
// Function to return the required count
int countNum(int a[], int n)
{
    int count = 0;
 
    for (int i = 0; i < n; i++) {
 
        // If a[i] or (a[i]+1) is a power of 2
        if (isPowerOfTwo(a[i]) || isPowerOfTwo(a[i] + 1))
            count++;
    }
 
    return count;
}
 
// Driver code
int main()
{
    int arr[] = { 5, 6, 9, 3, 1 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << countNum(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function that returns true if x is a power of 2
static boolean isPowerOfTwo(int x)
{
    if (x == 0)
        return false;
 
    // If x & (x-1) = 0 then x is a power of 2
    if ((x & (x - 1)) == 0)
        return true;
    else
        return false;
}
 
// Function to return the required count
static int countNum(int a[], int n)
{
    int count = 0;
 
    for (int i = 0; i < n; i++) {
 
        // If a[i] or (a[i]+1) is a power of 2
        if (isPowerOfTwo(a[i]) || isPowerOfTwo(a[i] + 1))
            count++;
    }
 
    return count;
}
 
// Driver code
public static void main(String args[])
{
    int arr[] = { 5, 6, 9, 3, 1 };
 
    int n = arr.length;
 
    System.out.println(countNum(arr, n));
 
}
}
 
// This code is contributed by
// Sahil_Shelangia

Python3

# Python 3 implementation of the approach
 
# Function that returns true if x
# is a power of 2
def isPowerOfTwo(x):
    if (x == 0):
        return False
 
    # If x & (x-1) = 0 then x is a power of 2
    if ((x & (x - 1)) == 0):
        return True
    else:
        return False
 
# Function to return the required count
def countNum(a, n):
    count = 0
 
    for i in range(0, n, 1):
         
        # If a[i] or (a[i]+1) is a power of 2
        if (isPowerOfTwo(a[i]) or
            isPowerOfTwo(a[i] + 1)):
            count += 1
 
    return count
 
# Driver code
if __name__ == '__main__':
    arr = [5, 6, 9, 3, 1]
 
    n = len(arr)
 
    print(countNum(arr, n))
 
# This code is contributed by
# Sanjit_Prasad

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function that returns true if x is a power of 2
static bool isPowerOfTwo(int x)
{
    if (x == 0)
        return false;
 
    // If x & (x-1) = 0 then x is a power of 2
    if ((x & (x - 1)) == 0)
        return true;
    else
        return false;
}
 
// Function to return the required count
static int countNum(int[] a, int n)
{
    int count = 0;
 
    for (int i = 0; i < n; i++) 
    {
        // If a[i] or (a[i]+1) is a power of 2
        if (isPowerOfTwo(a[i]) || isPowerOfTwo(a[i] + 1))
            count++;
    }
 
    return count;
}
 
// Driver code
public static void Main()
{
    int[] arr = { 5, 6, 9, 3, 1 };
    int n = arr.Length;
    Console.WriteLine(countNum(arr, n));
 
}
}
 
// This code is contributed by
// Mukul Singh

PHP

<?php
// PHP implementation of the approach
 
// Function that returns true if x is 
// a power of 2
function isPowerOfTwo( $x)
{
    if ($x == 0)
        return false;
 
    // If x & (x-1) = 0 then x is a 
    // power of 2
    if (!($x & ($x - 1)))
        return true;
    else
        return false;
}
 
// Function to return the required count
function countNum($a, $n)
{
    $cnt = 0;
 
    for ( $i = 0; $i < $n; $i++) 
    {
 
        // If a[i] or (a[i]+1) is a power of 2
        if (isPowerOfTwo($a[$i]) || 
            isPowerOfTwo($a[$i] + 1))
            $cnt++;
    }
 
    return $cnt;
}
 
// Driver Code
$arr = array( 5, 6, 9, 3, 1 );
 
$n = count($arr);
 
echo countNum($arr, $n);
 
// This code is contributed by 29AjayKumar
?>

Javascript

<script>
 
// Javascript implementation of the approach
     
// Function that returns true if x is a power of 2
function isPowerOfTwo(x)
{
    if (x == 0)
        return false;
   
    // If x & (x-1) = 0 then x is a power of 2
    if ((x & (x - 1)) == 0)
        return true;
    else
        return false;
}
 
// Function to return the required count
function countNum(a, n)
{
    let count = 0;
   
    for(let i = 0; i < n; i++)
    {
         
        // If a[i] or (a[i]+1) is a power of 2
        if (isPowerOfTwo(a[i]) || 
            isPowerOfTwo(a[i] + 1))
            count++;
    }
    return count;
}
 
// Driver code
let arr = [ 5, 6, 9, 3, 1 ];
let n = arr.length;
 
document.write(countNum(arr, n));
 
// This code is contributed by unknown2108
 
</script>

Output

Time Complexity: O(n), where n is the size of the given array.
Auxiliary Space: O(1)

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