869. 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 <= N <= 10^9

Solution:

class Solution:
    def reorderedPowerOf2(self, N: int) -> bool:
        c = Counter(str(N))
        # 10^9 = 2 ^30
        for i in range(30):
            n = int(math.pow(2, i))
            print()
            if Counter(str(n)) == c:
                return True
        return False

references

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

# leetcode challenge, medium, python

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869. Reordered Power of 2

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