<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 n = nums.size();
        int m = nums[0].size();
        vector<pair<int, int>>store;
        for (size_t k = 0; k < nums.size(); ++k)
            for (auto n: nums[k]) store.push_back({n, k});
        sort(store.begin(), store.end());
        int l = 0, r = 0;
        int mini = store[0].first;
        int maxi = store.back().first;
        unordered_map<int, int> mp;
        while(r < store.size()){
            mp[store[r].second]++;
            while(mp.size() >= n){
                if(store[r].first - store[l].first < maxi - mini){
                    maxi = store[r].first;
                    mini = store[l].first;
                }
                mp[store[l].second]--;
                if(mp[store[l].second] == 0)
                    mp.erase(store[l].second);
                l++;    
            }
            r++;
        }
        return {mini, maxi};
    }
};
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