new_contents: # Time:  O(nlogk)
  # Space: O(k)
  
  # You have k lists of sorted integers in ascending order.
  # Find the smallest range that includes at least one number from each of the k lists.
  #
  # We define the range [a,b] is smaller than range [c,d] if b-a < d-c or a < c if b-a == d-c.
  #
  # Example 1:
  # Input:[[4,10,15,24,26], [0,9,12,20], [5,18,22,30]]
  # Output: [20,24]
  # Explanation: 
  # 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].
  # Note:
  # The given list may contain duplicates, so ascending order means >= here.
  # 1 <= k <= 3500
  # -105 <= value of elements <= 10^5.
  # For Java users, please note that the input type has been changed to List<List<Integer>>.
  # And after you reset the code template, you'll see this point.
  
  class Solution(object):
      def smallestRange(self, nums):
          """
          :type nums: List[List[int]]
          :rtype: List[int]
          """
          left, right = float("inf"), float("-inf")
          min_heap = []
          for row in nums:
              left = min(left, row[0])
              right = max(right, row[0])
              it = iter(row)
              heapq.heappush(min_heap, (next(it, None), it))
          
          result = (left, right)
          while min_heap:
              (val, it) = heapq.heappop(min_heap)    
              val = next(it, None)
              if val is None:
                  break
              heapq.heappush(min_heap, (val, it))
              left, right = min_heap[0][0], max(right, val);
              if right - left < result[1] - result[0]:
                  result = (left, right)
          return result
  
subject: Create smallest-range.py