<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>>& a) {
            int n=a.size();
            int l=0,h=INT_MAX, minr=INT_MAX;
            int l1=0,h1=INT_MAX;
            multiset<pair<int,array<int,2>>> pq;

            for(int i=0;i<n;i++)
            {
                pq.insert({a[i][0],{i,0}});
            }
            l=(*pq.begin()).first,h=(*--pq.end()).first;
            if(h-l<minr)
            {
                minr=min(minr,h-l);
                h1=h;l1=l;
            }
        while(pq.size()==n)
        {
            pair<int,array<int,2>> pr=*pq.begin();
            pq.erase(pq.begin());
            if(pr.second[1]==a[pr.second[0]].size()-1)continue;
            pq.insert({a[pr.second[0]][pr.second[1]+1],{pr.second[0],pr.second[1]+1}});
            l=(*pq.begin()).first,h=(*--pq.end()).first;
            if(h-l<minr)
            {
                minr=min(minr,h-l);
                h1=h;l1=l;
            }
        }
        return {l1,h1};
    }
};
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