<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 minimumDeviation(vector<int>& nums) {
        vector<vector<int>> v;
        for(int i=0;i<nums.size();i++){
            vector<int> cur;
            if(nums[i]%2==0){
                int x=nums[i];
                while(x%2==0){
                    cur.push_back(x);
                    x=x/2;
                }
                if(x!=0){
                    cur.push_back(x);
                }
            }
            else{
                cur.push_back(nums[i]);
                cur.push_back(nums[i]*2);
            }
            sort(cur.begin(),cur.end());
            v.push_back(cur);
        }
        priority_queue<pair<int,pair<int,int>>,vector<pair<int,pair<int,int>>>,greater<pair<int,pair<int,int>>>> pq;
        int maxi=0;
        int n=nums.size();
        for(int i=0;i<n;i++){
            pq.push({v[i][0],{i,0}});
            maxi=max(maxi,v[i][0]);
        }
        int range=maxi-pq.top().first;
        while(true){
            auto it=pq.top();
            int mini=it.first;
            int row=it.second.first;
            int col=it.second.second;
            if(col==v[row].size()-1){
                break;
            }
            else{
                pq.pop();
                maxi=max(maxi,v[row][col+1]);
                pq.push({v[row][col+1],{row,col+1}});
                range=min(range,maxi-pq.top().first);
            }

        }
        return range;
        // for(auto it:v){
        //     for(auto at:it){
        //         cout<<at<<" ";
        //     }cout<<endl;
        // }
        return 0;
    }
};
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