1560. Most Visited Sector in a Circular Track

Problem

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

/**
 * @param {number} n
 * @param {number[]} rounds
 * @return {number[]}
 */
var mostVisited = function(n, rounds) {
    var start = rounds[0];
    var end = rounds[rounds.length - 1];
    if (end >= start) {
        return Array(end - start + 1).fill(0).reduce((arr, num, i) => {
            arr.push(start + i);
            return arr;
        }, []);
    } else {
        var arr1 = Array(n - start + 1).fill(0).reduce((arr, num, i) => {
            arr.push(start + i);
            return arr;
        }, []);
        var arr2 = Array(end).fill(0).reduce((arr, num, i) => {
            arr.push(i + 1);
            return arr;
        }, []);
        return arr2.concat(arr1);
    }
};

Explain:

if start <= end, return range[start, end]

else return range[0, end] + range[start, n]

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).