# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

 ` `

Output:

```2
```

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