<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) {
        priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> pq;
        for(int i=0;i<nums.size();i++)
        {
            for(int j=0;j<nums[i].size();j++)
            {
                pq.push({nums[i][j],i});
            }
        }
        vector<pair<int,int>> res;
        while(!pq.empty())
        {
            auto top=pq.top();
            pq.pop();
            res.push_back(top);
        }
        
        int i=0,j=0;
        map<int,int> m1;
        int mini=INT_MAX;
        int ti=-1;
        for(int i=0;i<nums.size();i++)
        m1[i]++;
        int cnt=nums.size();
        int rangemin=-1e5;
        int rangemax=1e5;
        while(j<res.size())
        {
            int b=res[j].second;
            if(m1.find(b)!=m1.end())
            {
                m1[b]--;
                if(m1[b]==0)
                cnt--;
                
            }
            if(cnt==0)
            {
                while(cnt==0)
                {
                    if(abs(res[j].first-res[i].first)<abs(rangemax-rangemin))
                    {
                        rangemin=res[i].first;
                        rangemax=res[j].first;
                    }
    
                    if(m1.find(res[i].second)!=m1.end())
                    {
                        m1[res[i].second]++;
                        if(m1[res[i].second]>0)
                        cnt++;
                    }
                    i++;
                }
            }
            j++;
        }
        vector<int> ans;
        ans.push_back(rangemin);
        ans.push_back(rangemax);
        return ans;
    }
};
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