<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]:
        #binsearch on each nums
        #o (n * klogn
        

        #Approach 1  -> do the merge first and somehow iterate through the new array to find the closest gap?
        # approach 2 -> do add all to heap ? then just move forward?
        
        mergeList = []
        for j in range(len(nums)):
            mergeList.extend([(nums[j][i], j) for i in range(len(nums[j]))])
        mergeList.sort()
    
        hashM = {i:0 for i in range(len(nums))}
        count = 0
        # heaq.heapify(heap)
        start = 0
        
        curAns = [mergeList[0][0], mergeList[-1][0]]

        curGap = curAns[1] - curAns[0]

        for end in range(len(mergeList)):
            curEle, ls = mergeList[end]
            if hashM[ls] == 0:
                count += 1
            hashM[ls] += 1
            while start < end:
                startEle, startGrp = mergeList[start]
                if hashM[startGrp] <= 1:
                    break
                else:
                    hashM[startGrp] -= 1
                    start += 1
            if count == len(nums) and curEle - mergeList[start][0] < curGap: #or (mergeList[end] - mergeList[start] < curGap)
                startEle, startGrp = mergeList[start]
                curAns = [startEle, curEle]
                curGap = curEle - startEle
        return curAns

            


            
            
       






24600