<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,pair<int,int>>, vector<pair<int,pair<int,int>>>, greater<pair<int,pair<int,int>>>>rec;
        int len = nums.size();
        for(int t=0;t<len;t++){
            rec.push({nums[t][0], {t, 0}});
        }
        vector<int>record(nums.size(), 0);
        queue<pair<int,int>>window;
        int count = 0;
        vector<int>ans;
        ans.push_back(-100000);
        ans.push_back(100000);
        while(!rec.empty()){
            int group = rec.top().second.first, spot = rec.top().second.second;
            record[group]++;
            if(record[group] == 1)count++;
            window.push({group, spot});
            while(record[window.front().first] > 1){
                record[window.front().first]--;
                window.pop();
            }
            if(count == nums.size() && nums[group][spot] - nums[window.front().first][window.front().second] < ans[1] - ans[0]){
                ans = {nums[window.front().first][window.front().second], nums[group][spot]};
            }

            
            if(spot < nums[group].size() - 1){
                rec.push({nums[group][spot + 1], {group, spot + 1}});

            }
            rec.pop();
        }
        return ans;
    }
};
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