# 869. Reordered Power of 2

https://leetcode.com/problems/reordered-power-of-2/

You are given an integer n. We reorder the digits in any order (including the original order) such that the leading digit is not zero. Return true if and only if we can do this so that the resulting number is a power of two.

Example 1: Input: n = 1 Output: true Example 2: Input: n = 10 Output: false

Constraints:

``````1 <= n <= 109
``````

• code
``````class Solution:
def reorderedPowerOf2(self, n: int) -> bool:
for i in itertools.permutations(str(n)):
if i != '0' and bin(int("".join(i))).count('1') == 1:
return True
return False
``````
• code
``````class Solution {
public boolean reorderedPowerOf2(int N) {
int[] A = count(N);
for (int i = 0; i < 31; ++i)
if (Arrays.equals(A, count(1 << i)))
return true;
return false;
}

// Returns the count of digits of N
// Eg. N = 112223334, returns [0,2,3,3,1,0,0,0,0,0]
public int[] count(int N) {
int[] ans = new int;
while (N > 0) {
ans[N % 10]++;
N /= 10;
}
return ans;
}
}
``````