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Most Visited Sector in a Circular Track - Array - Easy - LeetCode - MiniTV

Most Visited Sector in a Circular Track - Array - Easy - LeetCode - मिनी टीवी

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

public class Solution {
    public IList<int> MostVisited(int n, int[] rounds) {
        int start = rounds[0];
        int end = rounds[rounds.Length-1];
        
        var res = new List<int>();
        
        if(start <= end) {
            while(start<=end){
                res.Add(start++);
            }
        } else {
                int i=1;
                while(i<=end) {
                    res.Add(i++);
                }
                while(start <= n) {
                    res.Add(start++);
                }
        }
        return res;
    }
}

 


Time Complexity: O(n)

Space Complexity: O(n)