[leetcode] Reordered Power of 2

problem

code

public boolean isPowerOf2(int n) {
  	// ex let say 32 => 10000 & (10000 - 1 = 1111) => 000000 true
  	// 31 => 1111 & (1111 - 1 = 1110) =>  1110 false
	return x != 0 && (x & (x - 1) == 0);
}

2nd try: leetcode with bitwise operator

class Solution {
    public boolean reorderedPowerOf2(int N) {
        int[] A = count(N);
      	// 2^31 = 1073741824 > 10^9 = 1000000000
        for (int i = 0; i < 30; ++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[10];
        while (N > 0) {
            ans[N % 10]++;
            N /= 10;
        }
        return ans;
    }
}

Time: O(log^2N), count(1 << i) logN, equals LogN
Space: O(Log^N), why? there are 30 arrays for power of 2s < 10000000000