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: Simulation

Time complexity: O(m*n)
Space complexity: O(n)

class Solution {
public:
  vector<int> mostVisited(int n, vector<int>& rounds) {
    vector<int> counts(n);
    counts[rounds[0] - 1] = 1;
    for (int i = 1; i < rounds.size(); ++i)
      for (int s = rounds[i - 1]; ; ++s) {
        ++counts[s %= n];
        if (s == rounds[i] - 1) break;
      }
    const int max_count = *max_element(begin(counts), end(counts));
    vector<int> ans;
    for (int i = 0; i < n; ++i)      
      if (counts[i] == max_count) ans.push_back(i + 1);    
    return ans;
  }
};

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