# Reordered Power of 2 in C++

Suppose we have a positive integer N, we reorder the digits in any order (including the original order) such that the leading digit is non-zero. We have to check whether we can do this in a way such that the resulting number is a power of 2. So if the number is like 46, then the answer will be true.

To solve this, we will follow these steps −

• Define a method called count, this will take x as input

• ret := 0

• while x is not 0

• ret := ret + 10 ^ last digit of x

• x := x / 10

• return ret

• From the main method do the following −

• x := count(N)

• for i in range 0 to 31

• if count(2^i) = x, then return true

• return false

Let us see the following implementation to get better understanding −

## Example

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int count(int x){
int ret = 0;
while(x){
ret += pow(10, x % 10);
x /= 10;
}
return ret;
}
bool reorderedPowerOf2(int N) {
int x = count(N);
for(int i = 0; i < 32; i++){
if(count(1 << i) == x) return true;
}
return false;
}
};
main(){
Solution ob;
cout << (ob.reorderedPowerOf2(812));
}

### Input

812

## Output

1