<p>You are given an array <code>nums</code> of <code>n</code> positive integers.</p>

<p>You can perform two types of operations on any element of the array any number of times:</p>

<ul>
	<li>If the element is <strong>even</strong>, <strong>divide</strong> it by <code>2</code>.

	<ul>
		<li>For example, if the array is <code>[1,2,3,4]</code>, then you can do this operation on the last element, and the array will be <code>[1,2,3,<u>2</u>].</code></li>
	</ul>
	</li>
	<li>If the element is <strong>odd</strong>, <strong>multiply</strong> it by <code>2</code>.
	<ul>
		<li>For example, if the array is <code>[1,2,3,4]</code>, then you can do this operation on the first element, and the array will be <code>[<u>2</u>,2,3,4].</code></li>
	</ul>
	</li>
</ul>

<p>The <strong>deviation</strong> of the array is the <strong>maximum difference</strong> between any two elements in the array.</p>

<p>Return <em>the <strong>minimum deviation</strong> the array can have after performing some number of operations.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

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

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,10,8]
<strong>Output:</strong> 3
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>2 &lt;= n &lt;= 5 * 10<sup><span style="font-size: 10.8333px;">4</span></sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def minimumDeviation(self, nums: List[int]) -> int:
        nums = list(set(nums))

        import heapq
        minHeap, heapMax = [], 0
        # O(n * log max(nums))
        for n in nums:
            tmp = n
            while n % 2 == 0:
                n = n // 2
            minHeap.append((n, max(tmp, 2 * n)))
            heapMax = max(heapMax, n)

        # O(NlogM * logN)
        res = float("inf")
        heapq.heapify(minHeap)
        while len(minHeap) == len(nums):
            n, nMax = heapq.heappop(minHeap)
            res = min(res, heapMax - n)

            if n < nMax:
                heapq.heappush(minHeap, (n * 2, nMax))
                heapMax = max(heapMax, n * 2)

        return res


        # my sol
        # diff of any two
        # probably want to prioritize
        # minimizing diff between max and min
        # -> heap?

        # obs:
        # - you can only multiply an odd number by 2 once
        # - you can keep dividing by 2 until you hit odd
        # preprocess to start with all even array?

        # q:
        # what avg or num do elements converge to?

        def can_mult_by_two(num):
            return num % 2 == 0 and (num // 2) % 2 == 0

        for i in range(len(nums)):
            while can_mult_by_two(nums[i]):
                nums[i] //= 2

        maxval = max(nums)
        import heapq
        heapq.heapify(nums)


        while abs(maxval - nums[0]) > \
                abs(maxval - nums[0] * 2):
            heapq.heappush(nums, 2 * heapq.heappop(nums))

        
        return max(nums) - min(nums)




                

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