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

Solution(javascript)

/**
 * @param {number} n
 * @param {number[]} rounds
 * @return {number[]}
 */
const mostVisited = function (n, rounds) {
  const visited = {}

  let index = 0
  const start = rounds[0]
  let max = 0
  while (index < rounds.length) {
    for (let i = start; i <= n && index < rounds.length;) {
      visited[i] = (visited[i] || 0) + 1
      max = Math.max(visited[i], max)
      if (i === rounds[index]) {
        index += 1
      }
      if (i < n) {
        i += 1
      } else {
        i = 1
      }
    }
  }
  return Object.keys(visited)
    .filter(x => visited[x] === max).map(x => Number(x)).sort((a, b) => a - b)
}