# Solving Leetcode Interviews in Seconds with AI: Most Visited Sector in  a Circular Track


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1560" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's a breakdown of the optimal approach, complexity analysis, and the Python code:

*   **High-Level Approach:**
    *   Simulate the rounds, tracking the number of visits for each sector using a count array.
    *   Determine the maximum visit count.
    *   Collect all sectors with the maximum visit count and sort them in ascending order.

*   **Complexity Analysis:**
    *   Runtime Complexity: O(m + n), where m is the number of rounds and n is the number of sectors.
    *   Storage Complexity: O(n), for the count array.

	
	# Code
	```python
	def mostVisited(n: int, rounds: list[int]) -> list[int]:
    """
    Finds the most visited sectors in a circular track.

    Args:
        n: The number of sectors.
        rounds: A list of sectors visited in each round.

    Returns:
        A list of the most visited sectors sorted in ascending order.
    """

    counts = [0] * n
    start = rounds[0] - 1  # Adjust to 0-based indexing

    for i in range(1, len(rounds)):
        end = rounds[i] - 1  # Adjust to 0-based indexing

        if start <= end:
            for j in range(start, end + 1):
                counts[j] += 1
        else:
            for j in range(start, n):
                counts[j] += 1
            for j in range(0, end + 1):
                counts[j] += 1

        start = end

    max_visits = 0
    for count in counts:
        max_visits = max(max_visits, count)

    most_visited_sectors = []
    for i in range(n):
        if counts[i] == max_visits:
            most_visited_sectors.append(i + 1)  # Adjust back to 1-based indexing

    return most_visited_sectors
	```