Effectively fetch the smallest element in a heap of Python

To solve Leetcode #1675, I wrote the below code with the help of hints:

import sys
import heapq

class Solution:
    def minimumDeviation(self, nums: List[int]) -> int:
        n = len(nums)
        heapq.heapify(nums)
        _max = max(nums)
        ans = max(nums) - min(nums)
        while True:
            item = heapq.heappop(nums)
            if item % 2 == 1:
                heapq.heappush(nums, item * 2)
                _max = max(_max, item * 2)
                ans = min(ans, _max - heapq.nsmallest(1, nums)[0])
            else:
                heapq.heappush(nums, item)
                break
        print("stage1:", nums)
        nums = [-item for item in nums]
        heapq.heapify(nums)
        _max = max(nums)
        while True:
            item = heapq.heappop(nums)
            if item % 2 == 0:
                heapq.heappush(nums, item // 2)
                _max = max(_max, item // 2)
                ans = min(ans, _max - heapq.nsmallest(1, nums)[0])
            else:
                break
                
        return ans      

import sys

import heapq

​

class Solution:

    def minimumDeviation(self, nums: List[int]) -> int:

        n = len(nums)

        heapq.heapify(nums)

        _max = max(nums)

        ans = max(nums) - min(nums)

        while True:

            item = heapq.heappop(nums)

            if item % 2 == 1:

                heapq.heappush(nums, item * 2)

                _max = max(_max, item * 2)

                ans = min(ans, _max - heapq.nsmallest(1, nums)[0])

            else:

                heapq.heappush(nums, item)

                break

        print("stage1:", nums)

        nums = [-item for item in nums]

        heapq.heapify(nums)

        _max = max(nums)

        while True:

            item = heapq.heappop(nums)

            if item % 2 == 0:

                heapq.heappush(nums, item // 2)

                _max = max(_max, item // 2)

                ans = min(ans, _max - heapq.nsmallest(1, nums)[0])

            else:

                break

        return ans