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1560. Most Visited Sector in a Circular Track

Description

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

Solutions

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      public List<Integer> mostVisited(int n, int[] rounds) {
          int m = rounds.length - 1;
          List<Integer> ans = new ArrayList<>();
          if (rounds[0] <= rounds[m]) {
              for (int i = rounds[0]; i <= rounds[m]; ++i) {
                  ans.add(i);
              }
          } else {
              for (int i = 1; i <= rounds[m]; ++i) {
                  ans.add(i);
              }
              for (int i = rounds[0]; i <= n; ++i) {
                  ans.add(i);
              }
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      vector<int> mostVisited(int n, vector<int>& rounds) {
          int m = rounds.size() - 1;
          vector<int> ans;
          if (rounds[0] <= rounds[m]) {
              for (int i = rounds[0]; i <= rounds[m]; ++i) ans.push_back(i);
          } else {
              for (int i = 1; i <= rounds[m]; ++i) ans.push_back(i);
              for (int i = rounds[0]; i <= n; ++i) ans.push_back(i);
          }
          return ans;
      }
  };
  

• class Solution:
      def mostVisited(self, n: int, rounds: List[int]) -> List[int]:
          if rounds[0] <= rounds[-1]:
              return list(range(rounds[0], rounds[-1] + 1))
          return list(range(1, rounds[-1] + 1)) + list(range(rounds[0], n + 1))
  
  

• func mostVisited(n int, rounds []int) []int {
  	m := len(rounds) - 1
  	var ans []int
  	if rounds[0] <= rounds[m] {
  		for i := rounds[0]; i <= rounds[m]; i++ {
  			ans = append(ans, i)
  		}
  	} else {
  		for i := 1; i <= rounds[m]; i++ {
  			ans = append(ans, i)
  		}
  		for i := rounds[0]; i <= n; i++ {
  			ans = append(ans, i)
  		}
  	}
  	return ans
  }
  

• function mostVisited(n: number, rounds: number[]): number[] {
      const ans: number[] = [];
      const m = rounds.length - 1;
      if (rounds[0] <= rounds[m]) {
          for (let i = rounds[0]; i <= rounds[m]; ++i) {
              ans.push(i);
          }
      } else {
          for (let i = 1; i <= rounds[m]; ++i) {
              ans.push(i);
          }
          for (let i = rounds[0]; i <= n; ++i) {
              ans.push(i);
          }
      }
      return ans;
  }
  
  

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