<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>>v1;

        vector<vector<int>>v2;

        int k=nums.size();
        


        int i=0;
        for(auto a:nums){

            for(auto b:a){
               v1.push_back(make_pair(b,i));
            }
            i++;
        }

        sort(v1.begin(),v1.end());


        for(auto a:v1){
            cout<<a.first<<","<<a.second<<" ";
        }






        int j,m=INT_MAX,l,r;

        i=0;
        j=0;

        map<int,int>m1;

        while(j<size(v1)){
            m1[v1[j].second]++;

            if(m1.size()==k){
                while(m1.size()==k){
                    
                   if(v1[j].first-v1[i].first<m){
                    m=v1[j].first-v1[i].first;     // upadting on the basis of diff of elements not indexes 
                    l=i;
                    r=j;
                   }

                   m1[v1[i].second]--;
                   if(m1[v1[i].second]<=0) m1.erase(v1[i].second);
                   i++;
                }
            }

            j++;
        }


        



        



        return {v1[l].first,v1[r].first};
    }
};
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