1560. Most Visited Sector in a Circular Track
Description
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]
for0 <= i < m
Solution
most-visited-sector-in-a-circular-track.py
class Solution:
def mostVisited(self, n: int, rounds: List[int]) -> List[int]:
dic = {rounds[0]:1}
N = len(rounds)
for i in range(N-1):
this_round = rounds[i]
next_round = rounds[i+1]
while this_round != next_round:
if this_round < n:
this_round += 1
else:
this_round = 1
if this_round not in dic:
dic[this_round] = 1
else:
dic[this_round] += 1
m = max(dic.values())
res = []
for key in dic:
if dic[key] == m:
res.append(key)
res.sort()
return res
most-visited-sector-in-a-circular-track.cpp
class Solution {
public:
vector<int> mostVisited(int n, vector<int>& A) {
vector<int> res;
for (int i = A[0]; i <= A[A.size() - 1]; ++i)
res.push_back(i);
if (res.size() > 0) return res;
for (int i = 1; i <= A[A.size() - 1]; ++i)
res.push_back(i);
for (int i = A[0]; i <= n; ++i)
res.push_back(i);
return res;
}
};