Solution 1

Python3

class Solution:
    def reorderedPowerOf2(self, n: int) -> bool:
        def convert(n):
            cnt = [0] * 10
            while n:
                n, v = divmod(n, 10)
                cnt[v] += 1
            return cnt

        i, s = 1, convert(n)
        while i <= 10**9:
            if convert(i) == s:
                return True
            i <<= 1
        return False

Java

class Solution {
    public boolean reorderedPowerOf2(int n) {
        String s = convert(n);
        for (int i = 1; i <= Math.pow(10, 9); i <<= 1) {
            if (s.equals(convert(i))) {
                return true;
            }
        }
        return false;
    }

    private String convert(int n) {
        char[] cnt = new char[10];
        for (; n > 0; n /= 10) {
            cnt[n % 10]++;
        }
        return new String(cnt);
    }
}

C++

class Solution {
public:
    bool reorderedPowerOf2(int n) {
        vector<int> s = convert(n);
        for (int i = 1; i <= pow(10, 9); i <<= 1)
            if (s == convert(i))
                return true;
        return false;
    }

    vector<int> convert(int n) {
        vector<int> cnt(10);
        for (; n; n /= 10) ++cnt[n % 10];
        return cnt;
    }
};

Go

func reorderedPowerOf2(n int) bool {
    convert := func(n int) []byte {
        cnt := make([]byte, 10)
        for ; n > 0; n /= 10 {
            cnt[n%10]++
        }
        return cnt
    }
    s := convert(n)
    for i := 1; i <= 1e9; i <<= 1 {
        if bytes.Equal(s, convert(i)) {
            return true
        }
    }
    return false
}

Problem