<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>

#define ll long long int 
class Solution {
public:
    bool compare(ll l,ll r,ll lr,ll rr){
        if(r-l<rr-lr) return true;
        else if(r-l==rr-lr && l<lr) return true;
        return false;
    }
    vector<int> smallestRange(vector<vector<int>>& nums) {
        vector<pair<ll,ll>> vec;
        for(ll i=0;i<nums.size();i++){
            for(ll j=0;j<nums[i].size();j++){
                vec.push_back({nums[i][j],i});
            }
        }
        ll l=0,r=0;
        ll n=nums.size();
        int lr=-1e9,rr=1e9;
        map<ll,ll> mp;
        sort(vec.begin(),vec.end());
        while(r<vec.size()){
            mp[vec[r].second]++;
            while(l<=r && mp.size()==n){
                if(compare(vec[l].first,vec[r].first,lr,rr)){
                    lr=vec[l].first;
                    rr=vec[r].first;
                }
                mp[vec[l].second]--;
                if(mp[vec[l].second]==0) mp.erase(vec[l].second);
                l++;
            }
            r++;
        }
        vector<int> ans={lr,rr};
        return ans;
    }
};
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