<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 mn = INT_MAX;
        int mx = INT_MIN;

        auto cmp = [](const pair<int,int>& a, const pair<int,int>& b) { return a.first > b.first; };
        priority_queue<pair<int,int>, vector<pair<int,int>>, decltype(cmp)> pq(cmp);
        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>> items;
        while (!pq.empty()) {
            items.push_back(pq.top());
            pq.pop();
        }

        int l = 0;
        int r = 0;
        int dist = INT_MAX;
        int lb = INT_MIN;
        int ub = INT_MAX;
        unordered_map<int,int> lset;
        int k = nums.size();
        int total_have = 0;
        for (int l = 0; l < items.size(); l++) {
            while (total_have < k && r < items.size()) {
                auto [x, label] = items[r];
                if (lset[label] < 1) total_have++;
                lset[label]++;
                r++;
            }

            if (total_have < k) break;
            auto diff = abs(items[r-1].first - items[l].first);
            if (diff < dist) {
                dist = diff;
                lb = items[l].first;
                ub = items[r-1].first;
            }

            if (total_have >= k && l < items.size()) {
                auto [x, label] = items[l];
                if (lset[label] == 1) total_have--;
                lset[label]--;
            }
        }
        return {lb, ub};
    }
};
24708