<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<pair<int,int>>a;
       for(int i=0; i<nums.size(); i++){
        for(int j=0; j<nums[i].size(); j++){
            a.push_back({nums[i][j],i});
        }
       }
       sort(a.begin(),a.end());
       int mini=1e9; int diff=nums.size(); int mp[diff]; int total=0; int i=0;
       vector<int>ans;
       memset(mp,0,sizeof(mp));
       for(int j=0; j<a.size(); j++){
        mp[a[j].second]++;
        if(mp[a[j].second]==1)total++;
        if(total==diff){
            if(a[j].first-a[i].first+1<mini){
                mini=a[j].first-a[i].first+1;
                ans={a[i].first,a[j].first};
            }
        }
        
        while(total==diff){
            if(a[j].first-a[i].first+1<mini){
                mini=a[j].first-a[i].first+1;
                ans={a[i].first,a[j].first};
            }
            mp[a[i].second]--;
            if(mp[a[i].second]<=0)total--;
            i++; 
        }
       }
       return ans; 
    }
};
21278