<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:
    typedef pair<int, pair<int, int> > P;
    vector<int> smallestRange(vector<vector<int>>& nums) {
        priority_queue<P, vector<P> , greater<P> > pq;
        int mini = INT_MAX;
        int maxi = INT_MIN;
        int idx = 0;
        for(auto it:nums){
            mini = min(mini, it[0]);
            maxi = max(maxi, it[0]);
            pq.push({it[0], {idx, 0}});
            idx++;
        }

        int range = maxi - mini;
        vector<int> ans(2);
        ans[0] = mini;
        ans[1] = maxi;
        while(!pq.empty()){
            P front = pq.top();
            pq.pop();
            mini = front.first;
            int i = front.second.first;
            int j = front.second.second;
            int rangeCurr = maxi - mini;
            if(range > rangeCurr){
                range = rangeCurr;
                ans[0] = mini;
                ans[1] = maxi;
            }
            int size = nums[i].size();
            if(j+1 < size){
                maxi = max(maxi, nums[i][j+1]);
                pq.push({nums[i][j+1], {i, j+1}});
            }else{
                return ans;
            }
        }
        return ans;
    }
};
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