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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