<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>

import heapq

class Solution:
    def smallestRange(self, nums: List[List[int]]) -> List[int]:
        all_nums = []

        h = []
        for idx in range(len(nums)):
            heapq.heappush(h, (nums[idx][0], idx, 0))
        
        while h:
            top = heapq.heappop(h)
            val, idx_list, idx_element = top

            all_nums.append((val, idx_list))
            if idx_element < len(nums[idx_list]) - 1:
                heapq.heappush(h, (nums[idx_list][idx_element + 1], idx_list, idx_element + 1))
        
        curr_window = {}
        st, end = 0, 0
        best = float('inf')
        ret = (0, 0)

        while end < len(all_nums):
            new_element = all_nums[end]
            new_val, new_idx = new_element
            if new_idx not in curr_window:
                curr_window[new_idx] = 0
            curr_window[new_idx] += 1

            while len(curr_window) == len(nums):
                if all_nums[end][0] - all_nums[st][0] + 1 < best:
                    ret = (all_nums[st][0], all_nums[end][0])
                best = min(best, all_nums[end][0] - all_nums[st][0] + 1)

                curr_window[all_nums[st][1]] -= 1
                if curr_window[all_nums[st][1]] == 0:
                    del curr_window[all_nums[st][1]]
                st += 1
            
            end += 1
        
        return ret

            

24400