LeetCode 1560. Most Visited Sector in a Circular Track Solution in Java, C++, Python & More | Explanation + Code

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] for 0 <= i < m

Solutions

Solution 1: Considering the Relationship Between Start and End Positions

Since the end position of each stage is the start position of the next stage, and each stage is in a counterclockwise direction, we can determine the number of times each sector is passed based on the relationship between the start and end positions.

If rounds[0] ≤ rounds[m], then the sectors from rounds[0] to rounds[m] are passed the most times, and we can directly return all sectors within this interval.

Otherwise, the sectors from 1 to rounds[m] and the sectors from rounds[0] to n form the union of the most passed sectors, and we can return the union of these two intervals.

The time complexity is O(n), where n is the number of sectors. Ignoring the space consumption of the answer array, the space complexity is O(1).

PythonJavaC++GoTypeScript

class Solution: def mostVisited(self, n: int, rounds: List[int]) -> List[int]: if rounds[0] <= rounds[-1]: return list(range(rounds[0], rounds[-1] + 1)) return list(range(1, rounds[-1] + 1)) + list(range(rounds[0], n + 1))(code-box)

class Solution { public List<Integer> mostVisited(int n, int[] rounds) { int m = rounds.length - 1; List<Integer> ans = new ArrayList<>(); if (rounds[0] <= rounds[m]) { for (int i = rounds[0]; i <= rounds[m]; ++i) { ans.add(i); } } else { for (int i = 1; i <= rounds[m]; ++i) { ans.add(i); } for (int i = rounds[0]; i <= n; ++i) { ans.add(i); } } return ans; } }(code-box)

class Solution { public: vector<int> mostVisited(int n, vector<int>& rounds) { int m = rounds.size() - 1; vector<int> ans; if (rounds[0] <= rounds[m]) { for (int i = rounds[0]; i <= rounds[m]; ++i) { ans.push_back(i); } } else { for (int i = 1; i <= rounds[m]; ++i) { ans.push_back(i); } for (int i = rounds[0]; i <= n; ++i) { ans.push_back(i); } } return ans; } };(code-box)

func mostVisited(n int, rounds []int) []int { m := len(rounds) - 1 var ans []int if rounds[0] <= rounds[m] { for i := rounds[0]; i <= rounds[m]; i++ { ans = append(ans, i) } } else { for i := 1; i <= rounds[m]; i++ { ans = append(ans, i) } for i := rounds[0]; i <= n; i++ { ans = append(ans, i) } } return ans }(code-box)

function mostVisited(n: number, rounds: number[]): number[] { const ans: number[] = []; const m = rounds.length - 1; if (rounds[0] <= rounds[m]) { for (let i = rounds[0]; i <= rounds[m]; ++i) { ans.push(i); } } else { for (let i = 1; i <= rounds[m]; ++i) { ans.push(i); } for (let i = rounds[0]; i <= n; ++i) { ans.push(i); } } return ans; }(code-box)