Skip to content
Related Articles

Related Articles

Improve Article

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

  • Last Updated : 14 Apr, 2021

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: 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: arr[] = {2, 3, 7, 9, 15} 
Output:
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:
 

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: 
2

 




My Personal Notes arrow_drop_up
Recommended Articles
Page :