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We are given an array of positive integers. The goal is to find the count of numbers that can be made power of two by adding 1 to them utmost once.

We will check using log2(i) that number is power of two or can become power of two by adding 1 to it. If yes increment count.

Let’s understand with examples.

**Input** − arr[]= {1,3,2,5,6 },

**Output** − Count of numbers that can become power of 2: 3

**Explanation** − 1+1=2 → 21 , 3+1=4 → 22 , 2=21 others will become 5+1=6, 6+1=7

**Input** − arr[]= {2,4,8,16 },

**Output**− Count of numbers that can become power of 2: 4

**Explanation** − All 4 numbers are already power of 2.

We take an integer array arr[] initialized with random positive numbers.

Function powofTwo(int arr[],int n) takes an array and its length as input and returns a count of numbers that are or can be made power of two.

Take the initial count as 0.

Traverse array from i=0 to i<n.

For each element check if floor(log2(arr[i]))==ceil((log2(arr[i])) or floor(log2(arr[i]+1))==ceil((log2(arr[i]+1)) , if true increment count in both cases.

Return count as final result.

#include <bits/stdc++.h> #include <math.h> using namespace std; int powofTwo(int arr[],int n){ int count=0; for(int i=0;i<n;i++){ if( floor(log2(arr[i])) == ceil(log2(arr[i])) ) { count++; } else{ ++arr[i]; if( floor(log2(arr[i])) == ceil(log2(arr[i])) ) { count++; } } } return count; } int main(){ int Arr[]={ 5,6,9,3,1 }; int len=sizeof(Arr)/sizeof(Arr[0]); cout<<endl<<"Count of numbers with power of 2 possible: "<<powofTwo(Arr,len); return 0; }

If we run the above code it will generate the following output −

Count of numbers with power of 2 possible: 2

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