<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) {
    // Min-heap to store the elements of the current range
    // Each element is a tuple (value, row_index, col_index)
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> minHeap;

        int max_value = INT_MIN;      // Store the maximum value of the current range
        int start =0, end = INT_MAX;  // Variables to track the smallest range

    // Step 1: Initialize the heap by pushing the first element of each list
        for(int i =0; i<nums.size(); i++){
            minHeap.push({nums[i][0], i, 0});
            max_value = max(max_value, nums[i][0]);   // Track the maximum value in the current range
        }

    // Step 2: Process the heap until one of the lists is exhausted
        while(!minHeap.empty()){
            auto current  = minHeap.top();
            minHeap.pop();

            int min_Value = current[0]; // The minimum value in the current range
            int row = current[1];       // Row index (which list)
            int col = current[2];       // Column index (which element in the list)

    // Update the smallest range if the current range is smaller       
            if(max_value-min_Value < end-start){
                start = min_Value;
                end = max_value;
            }
    // Step 3: Move to the next element in the list
            if(col+1 < nums[row].size()){
                minHeap.push({nums[row][col+1], row, col+1});
                max_value = max(max_value, nums[row][col+1]); // Update the maximum value

            }
            else{
                break;
            }
        }
        return {start, end};
    }
};
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