<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) {
        vector<int> ans;
        ans.push_back(1);
        ans.push_back(INT_MAX);
        vector<pair<int,int>> arr;
        for(int i = 0;i<nums.size();i++){
            for(auto &j : nums[i]) arr.push_back({j,i});
        }
        sort(arr.begin(),arr.end());
        map<int,int> mp;
        for(int i = 0,j=0;i<arr.size();i++){
            mp[arr[i].second]++;
            
            if(mp.size()==nums.size()){
            while(mp.size() == nums.size()){
                mp[arr[j].second]--;
                if(mp[arr[j].second] == 0) mp.erase(arr[j].second);
                j++;
            }
            j--;
            mp[arr[j].second]++;

            if(arr[i].first-arr[j].first < ans[1]-ans[0]){
                ans[0] = arr[j].first;
                ans[1] = arr[i].first;
            }
            }
        }
        return ans;

    }
};
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