Minimize Deviation in Array
<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>
Greedy; Array; Ordered Set; Heap (Priority Queue)
class Solution {
public:
    int minimumDeviation(vector<int>& nums) {
        int mi = INT_MAX;
        priority_queue<int> pq;
        for (int v : nums) {
            if (v & 1) v <<= 1;
            pq.push(v);
            mi = min(mi, v);
        }
        int ans = pq.top() - mi;
        while (pq.top() % 2 == 0) {
            int x = pq.top() >> 1;
            pq.pop();
            pq.push(x);
            mi = min(mi, x);
            ans = min(ans, pq.top() - mi);
        }
        return ans;
    }
};