Minimize Deviation in Array Python Solution

lcpy gen -n 1675   # by problem number
lcpy gen -s minimize_deviation_in_array   # by problem name

Input: nums = [1,2,3,4]
Output: 1
Explanation: You can transform the array to [1,2,3,2], then to [2,2,3,2], then the deviation will be 3 - 2 = 1.

Input: nums = [4,1,5,20,3]
Output: 3
Explanation: You can transform the array after two operations to [4,2,5,5,3], then the deviation will be 5 - 2 = 3.

Input: nums = [2,10,8]
Output: 3

import heapq


class Solution:
    # Time: O(n * log(max(nums)) * log n)
    # Space: O(n)
    def minimum_deviation(self, nums: list[int]) -> int:
        # Raise every element to its largest reachable form (an odd x can only
        # grow once, to 2x); then repeatedly shrink the current max while it is
        # even, tracking the tightest window seen.
        heap: list[int] = []
        low = 1 << 62
        for num in nums:
            value = num * 2 if num % 2 else num
            heapq.heappush(heap, -value)
            low = min(low, value)

        best = 1 << 62
        while True:
            high = -heapq.heappop(heap)
            best = min(best, high - low)
            if high % 2:
                break
            half = high // 2
            low = min(low, half)
            heapq.heappush(heap, -half)
        return best