<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) {
        int n=nums.size();
        int range=0;
        for(int i=0;i<n;i++){
            range+=nums[i].size();
            // sort(nums[i].begin(),nums[i].end());
        }
        int res=INT_MAX;
        vector<int> ans={-1,-1};
        //now all sorted
        for(int i=0;i<range;i++){
            int smallest=nums[0][0];
            int ind=0;
            int largest=nums[0][0];
            for(int j=0;j<n;j++){
                // cout<<nums[j].size()<<" ";
                if(nums[j].size()==0) break;
                if(smallest>nums[j][0]){
                    smallest=nums[j][0];
                    ind=j;
                }
                largest=max(largest,nums[j][0]);
            }
            if(largest-smallest<res){
                res=largest-smallest;
                ans={smallest,largest};
            }
            // cout<<endl;
            nums[ind].erase(nums[ind].begin());
            if(nums[ind].size()==0) break;
        }

        return ans;
    }
};
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