1560. Most Visited Sector in a Circular Track

Problem Description

In this problem, we have a circular track with n sectors, each labeled from 1 to n. A marathon is taking place on this track, consisting of m rounds. Each round starts and ends at specified sectors, with the starting sector for each round given in an array rounds. Specifically, the ith round starts at sector rounds[i - 1] and ends at sector rounds[i]. The goal is to find out which sectors are the most visited throughout the marathon and return these sectors sorted in ascending order. It's important to notice that the sectors must be visited in the counter-clockwise direction and we wrap around the track each time we reach sector n.

Intuition

The solution to this problem relies on the observation that if the start sector of the marathon (rounds[0]) is less than or equal to the end sector (rounds[-1]), then all sectors in between are visited more frequently than any other sector. This is because each round always moves from a lower to a higher sector number, covering all sectors in between.

However, if the start sector is greater than the end sector, this indicates that the runners have passed the sector marked n and started a new lap, visiting the sectors from 1 to rounds[-1] again. Therefore, in this case, the most visited sectors will be from sector 1 to rounds[-1] and from rounds[0] to sector n.

The solution approach, therefore, is to check the relative positions of the start and end sectors and return the appropriate range of sectors:

1. If rounds[0] is less than or equal to rounds[-1], return a list of sectors ranging from rounds[0] to rounds[-1].
2. If rounds[0] is greater than rounds[-1], there are two ranges of sectors to consider. Combine the range from 1 to rounds[-1] with the range from rounds[0] to n into a single list and return it.

This insight leads to a straightforward solution that is intuitive and elegant, avoiding cumbersome calculations or tracking of the exact number of visits to each sector.

Solution Approach

The solution for finding the most visited sectors on the circular track follows a straightforward approach based on conditional logic rather than elaborate algorithms or data structures.

Here's a step-by-step explanation of the implementation:

1. Condition Check: First, we check if rounds[0], the starting sector of the first round, is less than or equal to rounds[-1], the ending sector of the last round.

   • If this condition is True, it means that the runners have not crossed the sector marked n from the last round to the first round, so we only need to consider the sectors that lie between rounds[0] and rounds[-1].

2. Range Construction (No Wrap-Around): When rounds[0] <= rounds[-1]:

   • We use the range function to generate a list starting from rounds[0] to rounds[-1], inclusive. This list represents the sectors that are covered in a direct path from the starting sector to the ending sector without any wrap-around.

3. Range Construction (With Wrap-Around): When rounds[0] > rounds[-1]:

   • This indicates that during the marathon, runners have wrapped around the track, passing by sector n and back to sector 1. We need to consider two ranges of sectors.
   • The first range is from sector 1 to rounds[-1], inclusive, which represents the sectors visited after the last wrap-around.
   • The second range is from rounds[0] to sector n, inclusive, which represents the sectors visited before the wrap-around occurs.
   • These two ranges are combined to form a complete list of the most visited sectors.

4. List Concatenation: To combine the two ranges when there is a wrap-around:

   • We utilize list concatenation to join the lists resulting from the two range calls into a single list, which is returned as the final result.

The code implementation uses only basic Python constructs such as conditional statements and list manipulations, thereby avoiding the need for complex algorithms or data structures. It leverages Python's range function to create lists directly corresponding to the ranges of sectors that are most visited, as deduced from the initial condition checks.

1class Solution:
2    def mostVisited(self, n: int, rounds: List[int]) -> List[int]:
3        if rounds[0] <= rounds[-1]:
4            return list(range(rounds[0], rounds[-1] + 1))
5        else:
6            return list(range(1, rounds[-1] + 1)) + list(range(rounds[0], n + 1))

The function mostVisited takes two parameters, n and rounds, which represent the number of sectors and the array of rounds, respectively. Within the function, we apply the logic outlined above to determine the most visited sectors and return them as a list.

The beauty of this solution lies in its simplicity and efficiency. Since it avoids additional computation by directly identifying the range of most visited sectors, it executes in constant time, making it suitable for even large values of n and numbers of rounds.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach using a circular track with 5 sectors (n = 5) and an array of rounds [1, 3, 5, 2]. We apply the steps from the solution approach to determine the most visited sectors.

Given:

• n = 5 (sectors are 1, 2, 3, 4, 5)
• rounds = [1, 3, 5, 2]

We start by applying the first step in our solution approach.

1. Condition Check: We compare rounds[0] to rounds[-1] (we check whether 1 <= 2).

   • In this case, the condition is True because 1 is less than 2.
   • Normally, this would imply that we only need to consider the sectors that lie between 1 and 2. However, since there is a wrap-around implied by the progression of sectors in the rounds from 5 back to 2, we need to consider the full marathon sequence.

2. Range Construction (With Wrap-Around): Since rounds[0] is not greater than rounds[-1] but there is a wrap-around from the third to the fourth round (from 5 back to 2), we consider the entire sequence of sectors visited.

   • The first range is from sector 1 to sector 5, inclusive.
   • The second range starts again at 1 after wrapping around and ends at 2.

3. List Concatenation: We combine the ranges from the first round and after the wrap-around:

   • The sectors visited are initially [1, 2, 3, 4, 5], and following the wrap-around, the sectors [1, 2] are visited again, leading to the sequence [1, 2, 3, 4, 5, 1, 2].
   • Since sectors 1 and 2 were visited in the last part of the marathon, they would be added to the most visited sectors list.

Hence, after evaluating the rounds in order, we determine that the most visited sectors would be [1, 2, 3, 4, 5], but [1, 2] have the highest visit count due to the wrap-around. We need to sort them to return the result as [1, 2].

Using the given solution code, the mostVisited function would thus return [1, 2, 3, 4, 5] + [1, 2], which simplifies to [1, 2, 3, 4, 5] since we are asked to return unique sectors visited. After sorting, the final answer remains [1, 2, 3, 4, 5], since all sectors are most visited due to the marathon sequence that was provided in the rounds array.

Solution Implementation

Time and Space Complexity

The time complexity of the code is primarily determined by the creation of the list that gets returned. This involves generating a range of integers, which can be done in constant time for each integer in the range, and then converting that range into a list.

• In the best-case scenario, where rounds[0] <= rounds[-1], we create a single range from rounds[0] to rounds[-1], which includes at most n elements. Thus, the time complexity for this case is O(n).

• In the worst-case scenario, where rounds[0] > rounds[-1], we create two ranges. The first is from 1 to rounds[-1] and the second is from rounds[0] to n. In the worst case, these two ranges can also include up to n elements combined, so the time complexity remains O(n).

The space complexity of the code is determined by the space needed to store the output list.

• In both cases mentioned above, the space complexity depends on the number of elements in the output list, which can be at most n. Thus, the space complexity for the algorithm is O(n).

Therefore, the final time complexity is O(n) and the space complexity is O(n).

Learn more about how to find time and space complexity quickly using problem constraints.