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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
Updated on: 19-Jul-2020
153 Views
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