Formatted question description: https://leetcode.ca/all/1560.html

1560. Most Visited Sector in a Circular Track (Easy)

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

Related Topics:
Array

Solution 1.

// OJ: https://leetcode.com/problems/most-visited-sector-in-a-circular-track/

// Time: O(NM)
// Space: O(N)
class Solution {
public:
vector<int> mostVisited(int n, vector<int>& A) {
vector<int> cnt(n + 1);
int from = A[0];
cnt[from]++;
for (int i = 1; i < A.size(); ++i) {
int to = A[i];
while (from != to) {
++from;
if (from > n) from = 1;
cnt[from]++;
}
}
vector<int> ans;
int mx = *max_element(begin(cnt), end(cnt));
for (int i = 1; i <= n; ++i) {
if (mx == cnt[i]) ans.push_back(i);
}
return ans;
}
};


Solution 2.

// OJ: https://leetcode.com/problems/most-visited-sector-in-a-circular-track/

// Time: O(N)
// Space: O(1)
// Ref: https://leetcode.com/problems/most-visited-sector-in-a-circular-track/discuss/806814/JavaC%2B%2BPython-From-Start-to-End
class Solution {
public:
vector<int> mostVisited(int n, vector<int>& A) {
vector<int> ans;
if (A[0] <= A.back()) {
for (int i = A[0]; i <= A.back(); ++i) ans.push_back(i);
} else {
for (int i = 1; i <= A.back(); ++i) ans.push_back(i);
for (int i = A[0]; i <= n; ++i) ans.push_back(i);
}
return ans;
}
};


Java

class Solution {
public List<Integer> mostVisited(int n, int[] rounds) {
int[] visits = new int[n];
visits[rounds[0] - 1]++;
int maxVisit = 1;
int roundsCount = rounds.length;
for (int i = 1; i < roundsCount; i++) {
int start = rounds[i - 1] - 1, end = rounds[i] - 1;
int curLength = (end - start + n) % n;
int index = start;
for (int j = 0; j < curLength; j++) {
index = (index + 1) % n;
visits[index]++;
maxVisit = Math.max(maxVisit, visits[index]);
}
}
List<Integer> list = new ArrayList<Integer>();
for (int i = 0; i < n; i++) {
if (visits[i] == maxVisit)