LeetCode-in-Java

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

Solution

import java.util.ArrayList;
import java.util.List;

public class Solution {
    public List<Integer> mostVisited(int n, int[] rounds) {
        List<Integer> res = new ArrayList<>();
        int start = rounds[0];
        int end = rounds[rounds.length - 1];
        int[] ans = new int[n + 1];
        while (start != end) {
            ans[start]++;
            start++;
            if (start > n) {
                start = 1;
            }
        }
        ans[end]++;
        for (int i = 1; i <= n; i++) {
            if (ans[i] != 0) {
                res.add(i);
            }
        }
        return res;
    }
}

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