<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:
struct compare{
    bool operator()(vector<int>& a, vector<int>& b){
        // min heap
        return a[0] > b[0];
    }
};
    vector<int> smallestRange(vector<vector<int>>& nums) {
        int ansEnd, ansStart;
        priority_queue<vector<int>, vector<vector<int>>, compare> pq;
        int maxi = INT_MIN;
        for(int i=0; i<nums.size(); i++){
            // fill first elements from each list
            pq.push({nums[i][0], i, 0});
            maxi = max(maxi, nums[i][0]);
        }      
        ansEnd = maxi;
        ansStart = pq.top()[0]; // since all the lists were sorted, top element of min heap is the smallest
        while(pq.size()){
            int x = pq.top()[0];
            int row = pq.top()[1];
            int col = pq.top()[2];
            pq.pop();
            if(maxi - x < ansEnd - ansStart){
                //shorter range is available
                ansStart = x;
                ansEnd = maxi;
            }
            if(col+1 < nums[row].size()){
                // take next element from the list from which x(minimum) is taken
                pq.push({nums[row][col+1], row, col+1});
                maxi = max(maxi, nums[row][col+1]);
            }
            else{
                break;
            }
        }
        return {ansStart, ansEnd};
    }
};
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