<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 k = nums.size();
        int range = INT_MAX;
        int start = -1;
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
        vector<int> m_idx(k);
        vector<int> m_val(k);
        vector<int> count(k, 0);
        unordered_set<int> s;
        map<int, int> cur_range;
        for (int i = 0; i < k; i++) {
            pq.push({nums[i][0], i});
            m_idx[i] = 1;
        }
        while(!pq.empty()) {
            auto [num, idx] = pq.top();
            // cout << num << " " << idx << endl;
            pq.pop();
            if(m_idx[idx] < nums[idx].size()) {
                pq.push(make_pair(nums[idx][m_idx[idx]], idx));
                m_idx[idx]++;
            }
            if (s.contains(idx)) {
                cur_range[m_val[idx]]--;
                if (cur_range[m_val[idx]] == 0) {
                    cur_range.erase(m_val[idx]);
                }
            } else {
                s.insert(idx);
            }
            cur_range[num]++;
            m_val[idx] = num;
            if (s.size() == k) {
                int r = cur_range.rbegin()->first - cur_range.begin()->first;
                if (r < range) {
                    range = r;
                    start = cur_range.begin()->first;
                }       
            }
        }
        return {start, start + range};
    }
};
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