text: """
  1560. Most Visited Sector in a Circular Track
  Easy
  
  Given an integer n and an integer array rounds. We have a circular track which consists of n sectors labeled from 1 to n. A marathon will be held on this track, the marathon consists of m rounds. The ith round starts at sector rounds[i - 1] and ends at sector rounds[i]. For example, round 1 starts at sector rounds[0] and ends at sector rounds[1]
  
  Return an array of the most visited sectors sorted in ascending order.
  
  Notice that you circulate the track in ascending order of sector numbers in the counter-clockwise direction (See the first example).
  
   
  
  Example 1:
  
  
  Input: n = 4, rounds = [1,3,1,2]
  Output: [1,2]
  Explanation: The marathon starts at sector 1. The order of the visited sectors is as follows:
  1 --> 2 --> 3 (end of round 1) --> 4 --> 1 (end of round 2) --> 2 (end of round 3 and the marathon)
  We can see that both sectors 1 and 2 are visited twice and they are the most visited sectors. Sectors 3 and 4 are visited only once.
  Example 2:
  
  Input: n = 2, rounds = [2,1,2,1,2,1,2,1,2]
  Output: [2]
  Example 3:
  
  Input: n = 7, rounds = [1,3,5,7]
  Output: [1,2,3,4,5,6,7]
   
  
  Constraints:
  
  2 <= n <= 100
  1 <= m <= 100
  rounds.length == m + 1
  1 <= rounds[i] <= n
  rounds[i] != rounds[i + 1] for 0 <= i < m
  """
  
  
  class Solution:
      def mostVisited(self, n: int, rounds: List[int]) -> List[int]:
          start, end = rounds[0], rounds[-1]
          return list(range(start, end + 1)) if start <= end else list(
              range(1, end + 1)) + list(range(start, n + 1))
  