<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) {
        if (nums.size() == 1) return {nums[0][0], nums[0][0]};
        vector<pair<int, int>> v, a, b;
        for (size_t i = 0; i < nums.size(); i++) {
            auto& x = nums[i];
            a.resize(x.size());
            for (size_t j = 0; j < x.size(); j++) a[j] = {x[j], i};
            b.resize(v.size() + a.size());
            merge(v.begin(), v.end(), a.begin(), a.end(), b.begin());
            swap(v, b);
        }
        vector<int> idx(nums.size(), -1);
        int cnt = 0;
        deque<pair<int, int>> q;
        int mn = INT_MAX, l = 0, r = 0;
        for (size_t i = 0; i < v.size(); i++) {
            auto [val, g] = v[i];
            if (idx[g] == -1) cnt++;
            idx[g] = i;
            q.emplace_back(g, i);
            if (cnt == nums.size()) {
                while (!q.empty()) {
                    auto [g, j] = q.front();
                    if (idx[g] == j) break;
                    q.pop_front();
                }
                auto [g, j] = q.front();
                q.pop_front();
                int dif = val - v[j].first;
                if (dif < mn) mn = dif, l = v[j].first, r = val;
                idx[g] = -1;
                cnt--;
            }
        }
        return {l, r};
    }
};
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