Reordered Power of 2

Starting with a positive 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 in a way such that the resulting number is a power of 2.

Example 1:

Input: 1
Output: true

Example 2:

Input: 10
Output: false

Example 3:

Input: 16
Output: true

Example 4:

Input: 24
Output: false

Example 5:

Input: 46
Output: true

Note:

1. 1 <= N <= 10^9

Solution:

TC - O(log^2 n)

public class Solution {
    public boolean reorderedPowerOf2(int N) {
        int[] digits = new int[10];
        for (int i = N; i > 0; i /= 10) 
        {
            digits[i % 10]++;
        }
        int num = 1;
        int bits = 1;
        while (bits < 32) {
            if (isDigitsMatch(num, digits)) {
                return true;
            }
            num =num*2;
            ++bits;
        }
        return false;
    }

    private boolean isDigitsMatch(int n, int[] digits) 
    {        
        int[] temp = new int[10];
        for (int i = n; i > 0; i /= 10) 
        {
            temp[i % 10]++;
        }
        int j=0;
        while(j<10)
        {
            if(temp[j]!=digits[j]) return false;
            j++;
        }
        return true;
    }

}