1560. Most Visited Sector in a Circular Track.cpp

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 {
public:
    vector<int> mostVisited(int n, vector<int>& rounds) {
        vector<int> res;
        vector<int> sector(n, 0);
        int len=rounds.size();
        for(int i=1;i<len;i++){
            int start=rounds[i-1];
            int end=rounds[i];
            if(start<=end){
                for(int j=start;j<end;j++)
                    sector[j]++;
            } else {
                for(int j=start; j<n;j++)
                    sector[j]++;
                for(int j=0;j<end;j++)
                    sector[j]++;
            }
        }
        sector[rounds[0]-1]++; // for initial start
        int maxi=sector[0];
        for(int i=1;i<n;i++){
            if(sector[i]>maxi)
                maxi=sector[i];
        }
        for(int i=0;i<n;i++){
            if(sector[i]==maxi)
                res.push_back(i+1);    // we are considering 0 as round 1 and n-1 as round n
        }
        return res;
    }
};