Smallest Range Covering Elements from K Lists
<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>
Greedy; Array; Hash Table; Sorting; Sliding Window; Heap (Priority Queue)
class Solution {
public:
    vector<int> smallestRange(vector<vector<int>>& nums) {
        int n = 0;
        for (auto& v : nums) n += v.size();
        vector<pair<int, int>> t(n);
        int k = nums.size();
        for (int i = 0, j = 0; i < k; ++i) {
            for (int v : nums[i]) {
                t[j++] = {v, i};
            }
        }
        sort(t.begin(), t.end());
        int j = 0;
        unordered_map<int, int> cnt;
        vector<int> ans = {-1000000, 1000000};
        for (int i = 0; i < n; ++i) {
            int b = t[i].first;
            int v = t[i].second;
            ++cnt[v];
            while (cnt.size() == k) {
                int a = t[j].first;
                int w = t[j].second;
                int x = b - a - (ans[1] - ans[0]);
                if (x < 0 || (x == 0 && a < ans[0])) {
                    ans[0] = a;
                    ans[1] = b;
                }
                if (--cnt[w] == 0) {
                    cnt.erase(w);
                }
                ++j;
            }
        }
        return ans;
    }
};