<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 compare{
  public:
  bool operator()(pair<int,pair<int,int>>a ,pair<int,pair<int,int>>b){
      return a.first>b.first;
  }  
};

class Solution {
public:
    vector<int> smallestRange(vector<vector<int>>& nums) {
       priority_queue<pair<int,pair<int,int>>, vector<pair<int,pair<int,int>>>, compare> pq;

       int maxi= INT_MIN;
       for(int i=0; i<nums.size(); i++){
        pair<int,pair<int,int>> p= make_pair(nums[i][0], make_pair(i,0));
        pq.push(p);
        maxi= max(maxi,p.first);
       }

       int ansStart= pq.top().first;
       int ansEnd= maxi; 

       while(!pq.empty()){
        int topData= pq.top().first;
        pair<int,int> coordinates= pq.top().second;
        int row= coordinates.first;
        int col= coordinates.second;
        pq.pop();

        if((maxi-topData)<(ansEnd-ansStart)){
            ansStart= topData;
            ansEnd= maxi;
        }
        if(col+1<nums[row].size()){
           pair<int,pair<int,int>> p= make_pair(nums[row][col+1], make_pair(row,col+1));
           pq.push(p);
           maxi= max(maxi,p.first);
        
        }
        else{
            break;
        }
       }
       vector<int> ans;
       ans.push_back(ansStart);
       ans.push_back(ansEnd);
       return ans;
    }
};
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