<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:

    def smallestRange(self, nums: List[List[int]]) -> List[int]:
        flattened = [[x, idx] for idx, vals in enumerate(nums) for x in vals]
        flattened.sort()
        count = Counter()
        pointer = 0
        res = [float('-inf'), float('inf')]

        for right_value, origin_list_index  in flattened:
            count[origin_list_index ] += 1
            while len(count) == len(nums):
                left_value = flattened[pointer][0]
                size_diff = right_value - left_value - (res[1] - res[0])
                if size_diff < 0 or (size_diff == 0 and left_value < res[0]):
                    res = [left_value, right_value]
                left_origin_index = flattened[pointer][1]
                count[left_origin_index  ] -= 1
                if count[left_origin_index  ] == 0:
                    count.pop(left_origin_index  )
                pointer += 1
        return res

24300