<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) {
        int minimum,maximum=INT_MIN;
        vector<pair<int,pair<int,int>>>temp1;
        for(int i=0;i<nums.size();i++){
            temp1.push_back(make_pair(nums[i][0],make_pair(i,0)));
            maximum=max(maximum,nums[i][0]);
        }
        priority_queue<pair<int,pair<int, int>>,vector<pair<int,pair<int,int>>>,greater<pair<int,pair<int, int>>>>p(temp1.begin(),temp1.end());
        minimum=p.top().first;
        vector<int>ans(2);
        ans[0]=minimum;
        ans[1]=maximum;
        pair<int,pair<int,int>>temp;
        int col,row;
        while(p.size()==nums.size()){
        temp=p.top();
        p.pop();
        row=temp.second.first;
        col=temp.second.second;
        if(col+1<nums[row].size()){
            col++;
            p.push(make_pair(nums[row][col],make_pair(row,col)));
            maximum=max(maximum,nums[row][col]);
            minimum=p.top().first;
            if(maximum-minimum<ans[1]-ans[0]){
               ans[0] =minimum;
               ans[1]=maximum;
            }
        }
        }
        return ans;

    }
};
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