<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 {
public:
    int eo(vector<int>& nums){
        priority_queue<int> pq; 
        priority_queue<int, vector<int>, greater<int>> mh;
        for(int i = 0; i < nums.size(); i++) {
            pq.push(nums[i]);
            mh.push(nums[i]);
        } 
        int mn = pq.top() - mh.top();
        
        while (pq.top() % 2 == 0) {
            int k = pq.top();
            pq.pop();
            pq.push(k / 2);
            mh.push(k / 2);
            if (mn < pq.top() - mh.top()) break;
            mn = min(mn, pq.top() - mh.top());
        }
        
        while (mh.top() % 2) {
            int k = mh.top();
            mh.pop();
            mh.push(k * 2);
            pq.push(k * 2);
            if (mn < pq.top() - mh.top()) break;
            mn = min(mn, pq.top() - mh.top());
        }
        return mn;
    }

    int oe(vector<int>& nums){
        priority_queue<int> pq; 
        priority_queue<int, vector<int>, greater<int>> mh;
        for(int i = 0; i < nums.size(); i++) {
            pq.push(nums[i]);
            mh.push(nums[i]);
        } 
        int mn = pq.top() - mh.top();
        while (mh.top() % 2) {
            int k = mh.top();
            mh.pop();
            mh.push(k * 2);
            pq.push(k * 2);
            // if (mn < pq.top() - mh.top()) break;
            mn = min(mn, pq.top() - mh.top());
        }
        while (pq.top() % 2 == 0) {
            int k = pq.top();
            pq.pop();
            pq.push(k / 2);
            mh.push(k / 2);
            // if (mn < pq.top() - mh.top()) break;
            mn = min(mn, pq.top() - mh.top());
        }
        return mn;
    }

    // int gr(vector<int>& nums){
    //     priority_queue<int> pq; 
    //     priority_queue<int, vector<int>, greater<int>> mh;
    //     for(int i = 0; i < nums.size(); i++) {
    //         pq.push(nums[i]);
    //         mh.push(nums[i]);
    //     } 
    //     int mn = pq.top() - mh.top();
    //     while(1){
    //         int k= pq.top();
    //         int l = mh.top();
    //         if(k%2==0 && l%2){

    //         }
    //         else if(l%2){
    //             int k = mh.top();
    //             mh.pop();
    //             mh.push(k * 2);
    //             pq.push(k * 2);
    //             if (mn < pq.top() - mh.top()) break;
    //             mn = min(mn, pq.top() - mh.top())
    //         }
    //         else if(k%2==0{
                
    //         }
    //         else break;
    //     }
    //     return mn;
    // }

    int minimumDeviation(vector<int>& nums) {
        return min(oe(nums), eo(nums));
    }
};

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