<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)
        {
           	//storing as (data,(row,col)) in pair
           	// pair<int, pair<int, int>> p;

            priority_queue<pair<int, pair<int, int>>, vector< pair<int, pair<int, int>>>, greater<pair<int, pair<int, int>>>> pq;

            int mini = INT_MAX;
            int maxi = INT_MIN;

           	//Inserting frist element of each row into heap
           	//this loop will take klogk
            for (int i = 0; i < nums.size(); i++)
            {
                pq.push(make_pair(nums[i][0], make_pair(i, 0)));
                maxi = max(maxi, nums[i][0]);
            }

            mini = pq.top().first;
            vector<int> ans(2);
            ans[0] = mini;
            ans[1] = maxi;
            pair<int, pair<int, int>> temp;
            int row, col, elm;

            while (pq.size() == nums.size())
            {
                temp = pq.top();
                pq.pop();
                elm = temp.first;
                row = temp.second.first;
                col = temp.second.second;

                //if next elem of top exsit we will push it in pq
                if(col+1 <nums[row].size()){
                    col++;
                    pq.push(make_pair(nums[row][col],make_pair(row,col)));
                    maxi=max(maxi,nums[row][col]);
                    mini=pq.top().first;

                    //if we have got smaller range them update asn
                    if(maxi-mini<ans[1]-ans[0]){
                        ans[0]=mini;
                        ans[1]=maxi;
                    }
                }
            }

            return ans;
        }
};
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