<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>

const int inf = 1e9 ;
class Solution {
public:
    void update(vector<int>&ans,
        priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>>&pq,
        int maxi){
            int mn = (pq.top()).first ;
            if(ans[1]-ans[0]>maxi-mn){
                ans = {mn,maxi} ;
            }
            else if(ans[1]-ans[0] == maxi-mn){
                ans = min(ans,{mn,maxi}) ;
            }
        }
    vector<int> smallestRange(vector<vector<int>>& nums) {

        priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>>pq ;
        int maxi = -inf ;
        int n = nums.size() ;
        for(int i =0;i<n;i++){
            pq.push({nums[i][0],i}) ;
            maxi = max(maxi,nums[i][0]) ;
        }
        vector<int> ans = {0,inf} ;
        vector<int>curr(n,0) ;
        update(ans,pq,maxi) ;
        while(true){
            pair<int,int> p = pq.top() ;
            pq.pop() ;
            int i = p.second ;
            curr[i]++ ;
            if(curr[i]==nums[i].size())break;
            pq.push({nums[i][curr[i]],i}) ;
            maxi = max(maxi,nums[i][curr[i]]) ;
            update(ans,pq,maxi) ;

        }
        return ans ;


    }
};
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