<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 Info{
    public :
        int data;
        int rowIndex;
        int colIndex;

    Info(int val , int row, int col){
        this->data = val;
        this->rowIndex = row;
        this->colIndex = col; 
    }
};

class compare{
    public: 
       
        bool operator()(Info* a,Info* b) {
            return a->data > b->data ;
        }
};

class Solution {
public:
    vector<int> smallestRange(vector<vector<int>>& nums) {
        //Min heap
        priority_queue< Info *, vector<Info*> , compare > pq;

        int maxi = INT_MIN;
        int mini = INT_MAX;

        //process first k elements 
        for(int i=0;i<nums.size();i++){
            int element = nums[i][0];
            int row = i;
            int col = 0;
            Info * temp = new Info(element,row,col);
            pq.push(temp);

            mini = min(mini,element);
            maxi = max(maxi,element);
        }

        int ansStart = mini;
        int ansEnd = maxi;

        //process karenge ab
        while(!pq.empty()){
            Info* topNode = pq.top();
            int topData = topNode->data;
            int topCol = topNode->colIndex;
            int topRow = topNode->rowIndex;
            pq.pop();

            mini = topNode->data;

            //maxi ki value pehle se padi hui hai
            //compare karlete hai for smaller range
            if((maxi - mini) < (ansEnd - ansStart)){
                //update ans ;
                ansStart = mini;
                ansEnd = maxi;
            }

            //insertion vala step
            if(topCol + 1 < nums[topRow].size()){
                int newElement = nums[topRow][topCol+1];
                maxi = max(maxi, newElement);
                Info* newInfo = new Info(newElement,topRow,topCol+1);
                pq.push(newInfo);
            }
            else{
                //aage koi element nahi h push karne ko
                //mean to say ki agar ak bhi array empty jo jae to break kar denge loop  
                break;
            }

            //Memory leak avoiding 
            delete topNode;

        }

        vector<int>ans;
        ans.push_back(ansStart);
        ans.push_back(ansEnd);

        return ans;
    }
};
23238