<p>You have <code>k</code> lists of sorted integers in <strong>non-decreasing&nbsp;order</strong>. Find the <b>smallest</b> range that includes at least one number from each of the <code>k</code> lists.</p>

<p>We define the range <code>[a, b]</code> is smaller than range <code>[c, d]</code> if <code>b - a &lt; d - c</code> <strong>or</strong> <code>a &lt; c</code> if <code>b - a == d - c</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [[4,10,15,24,26],[0,9,12,20],[5,18,22,30]]
<strong>Output:</strong> [20,24]
<strong>Explanation: </strong>
List 1: [4, 10, 15, 24,26], 24 is in range [20,24].
List 2: [0, 9, 12, 20], 20 is in range [20,24].
List 3: [5, 18, 22, 30], 22 is in range [20,24].
</pre>

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

<pre>
<strong>Input:</strong> nums = [[1,2,3],[1,2,3],[1,2,3]]
<strong>Output:</strong> [1,1]
</pre>

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

<ul>
	<li><code>nums.length == k</code></li>
	<li><code>1 &lt;= k &lt;= 3500</code></li>
	<li><code>1 &lt;= nums[i].length &lt;= 50</code></li>
	<li><code>-10<sup>5</sup> &lt;= nums[i][j] &lt;= 10<sup>5</sup></code></li>
	<li><code>nums[i]</code>&nbsp;is sorted in <strong>non-decreasing</strong> order.</li>
</ul>

class Solution {
public:
    vector<int> smallestRange(vector<vector<int>>& nums) {
        vector<int> ans(2,0);
        int k=nums.size();
        int mn=INT_MAX;
        int mx=INT_MIN;
        priority_queue<vector<int>,vector<vector<int>>,greater<vector<int>>> pq;
        for(int i=0;i<k;i++){
            pq.push({nums[i][0],i,0});
            mn=min(mn,nums[i][0]);
            mx=max(mx,nums[i][0]);
        }
        ans[0]=mn;
        ans[1]=mx;
        while(true){
            auto temp=pq.top();
            pq.pop();
            int i=temp[1];
            int j=temp[2];
            j++;
            if(j<nums[i].size()){
                int num2=nums[i][j];
                mx=max(mx,num2);
                pq.push({num2,i,j});
            }else return ans;
            mn=pq.top()[0];
            if((mx-mn)<(ans[1]-ans[0])){
                ans[0]=mn;
                ans[1]=mx;
            }
        }
        return ans;
    }
};
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