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

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output:

2


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