<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) {
        multiset<pair<int,int>>s;
        int n = nums.size();
        int mnLen = INT_MAX;
        int l=-1,r=-1;
        for(int i=0;i<n;i++){
            s.insert(make_pair(nums[i].back(),i));
            nums[i].pop_back();
        }
        while(!s.empty()){
            int len = s.rbegin()->first - s.begin()->first;
            if(len<mnLen){
                mnLen = len;
                l = s.begin()->first;
                r = s.rbegin()->first;
            }
            else if(len == mnLen && s.begin()->first<l){
                l = s.begin()->first;
                r = s.rbegin()->first;
            }

            int idx = s.rbegin()->second;
            s.erase(--s.end());
            if(!nums[idx].empty()){
                s.insert({nums[idx].back(),idx});
                nums[idx].pop_back();
            }else break;
        }
        return {l,r};
    }
};
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