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# 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:

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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 ``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);` `    ``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

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## Javascript

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Output:
`2`

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