algoadvance

You are given an integer n and an integer array rounds. There is a circular track which consists of n sectors labeled from 1 to n. A marathon runner starts at sector rounds[0] and runs in the clockwise direction sequentially through rounds. The runner finishes at the position rounds[rounds.length - 1]. The task is to find the most visited sectors in the track. You need to return these sectors sorted in ascending order.

Clarifying Questions

1. Are there multiple runs provided in the rounds array, or is it just one continuous run?
   • Answer: The rounds array represents one continuous run.
2. Is it guaranteed that the rounds array will have at least two elements, i.e., a start and an end sector?
   • Answer: Yes, it is guaranteed as per the problem statement.
3. Can the length of rounds be greater than the number of sectors n?
   • Answer: Yes, because the runner may circle around the track multiple times.

Strategy

1. Initialization: We’ll keep track of which sectors are visited the most.
2. Simulation: As the runner starts from the first sector in rounds and sequentially goes through the track according to the given array, we will simulate this movement and count visits to each sector.
3. Result Extraction: We’ll then determine the most visited sectors, sort them, and return the result.

Given that the track is circular:

• If rounds[i] is less than rounds[i-1], it implies that the runner passes the sector n and starts from 1 again.

Steps

1. Initialize the starting sector.
2. Iterate over the rounds array to simulate the track traversal and count sector visits.
3. Sort and return the most visited sectors.

Code

def most_visited(n, rounds):
    # Initialize sector visit counts
    visit_counts = [0] * (n + 1)
    
    # Track sectors visited during each round
    start = rounds[0]
    for end in rounds:
        if start <= end:
            for sector in range(start, end + 1):
                visit_counts[sector] += 1
        else:
            for sector in range(start, n + 1):
                visit_counts[sector] += 1
            for sector in range(1, end + 1):
                visit_counts[sector] += 1
        start = end
    
    # Find maximum visit count
    max_visit = max(visit_counts)
    
    # Extract sectors with the maximum visit count
    result = [i for i in range(1, n + 1) if visit_counts[i] == max_visit]
    
    return sorted(result)

# Example usage
n = 4
rounds = [1, 3, 1, 2]
print(most_visited(n, rounds))  # Output: [1, 2]

Time Complexity

• Traversal of Round Array: We traverse each element of the rounds array once, iterating through sectors between two points in each step which is dominated by n, resulting in O(m + n), where m is the length of rounds.
• Finding the Maximum and Extracting Results: This involves finding the maximum from n sectors and extracting sectors, which is O(n).

Therefore, the overall time complexity is O(m + n).

The space complexity is O(n) for the visit_counts array.

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