900 Reordered Power Of 2

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

class Solution:
    def reorderedPowerOf2(self, n: int) -> bool:
        l = list(str(n))
        res = []
        l.sort(key= lambda x : -int(x))
        print(l)
        k = 1
        while(k <= int("".join(l)) ):
            b = list(str(k))
            b.sort(key= lambda x : -int(x))
            if "".join(b) == "".join(l):
                return True
            k *= 2
        return False



        ln = len(str(n))

        def recurse(l):
            if len(l) == 0:
                return []
            if len(l) == 2:
                return [l[0] + l[1], l[1] + l[0]]
            if len(l) == 1:
                return [l[0]]
            res = set()
            for i,k in enumerate(l):
                a = recurse(l[:i])
                b = recurse(l[i+1:])
                print(a,b)
                if len(a) == 0:
                    a.append("")
                if len(b) == 0:
                    b.append("")
                for m in a:
                    for n1 in b:
                        res.add(m + l[i] + n1)
                for m in b:
                    for n1 in a:
                        res.add(m + l[i] + n1)
            print(res)
            return res
        for k in recurse(l):

            if  ln == len(str(int(k))):
                print(k)
                return True
        return False