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Leetcode 1560. Most Visited Sector in a Circular Track Sure! Let’s tackle the problem step by step.

Problem Statement

You are given an integer n and an integer array rounds. The track consists of n sectors numbered from 1 to n. A marathoner will be running on the track starting at sector rounds[0] and will run in the direction specified by the subsequent elements in the array rounds.

Return an array of the most visited sectors sorted in ascending order.

Clarifying Questions

1. Will the elements of the rounds array be non-decreasing?
2. Can the starting sector and the ending sector of the marathon overlap to the next or previous lap?
3. Are all input values within constraints that fit within the typical bounds of integer computations?

Strategy

The strategy to solve this problem is as follows:

1. Initialization:
   • Start at rounds[0].
2. Traversal and Sector Counting:
   • Traverse through the rounds array and for each segment count the sectors covered.
   • Keep track of the frequency of visits to each sector during the traversal.
3. Determine the Most Visited Sectors:
   • Find the maximum frequency of visits across all sectors.
   • Collect all sectors which have this maximum frequency.
4. Output Result:
   • Sort the collected sectors in ascending order to meet the requirement.

Code

Here is how you can implement this strategy in C++:

#include <vector>
#include <algorithm>
#include <unordered_map>
#include <iostream>

std::vector<int> mostVisited(int n, std::vector<int>& rounds) {
    // Create a frequency map to count visits to each sector
    std::vector<int> visitCount(n + 1, 0);
    
    int start = rounds[0];
    visitCount[start]++;
    
    for (int i = 1; i < rounds.size(); ++i) {
        int current = start;
        int next = rounds[i];
        
        if (next > current) {
            for (int sector = current + 1; sector <= next; ++sector) {
                visitCount[sector]++;
            }
        } else if (next < current) {
            for (int sector = current + 1; sector <= n; ++sector) {
                visitCount[sector]++;
            }
            for (int sector = 1; sector <= next; ++sector) {
                visitCount[sector]++;
            }
        }
        start = next;
    }

    // Find the maximum visit count
    int maxVisit = *std::max_element(visitCount.begin(), visitCount.end());
    
    // Collect all sectors with the maximum visit count
    std::vector<int> result;
    for (int sector = 1; sector <= n; ++sector) {
        if (visitCount[sector] == maxVisit) {
            result.push_back(sector);
        }
    }
    
    return result;
}

int main() {
    int n = 4;
    std::vector<int> rounds = {1, 3, 1, 2};
    std::vector<int> result = mostVisited(n, rounds);
    for (int sector : result) {
        std::cout << sector << " ";
    }
    std::cout << std::endl;  // Output: 1 2
    return 0;
}

Time Complexity

• Initialization: O(n) for initializing the visitCount vector.
• Traversal and Counting: O(m) where m is the size of the rounds array. However, there could be an internal loop that runs in worst case O(n) for each segment.
  • Summing up, the complexity would be O(n * m) in the worst case.
• Maximum Visit Determination: O(n) to find the maximum visit count.
• Result Collection: O(n) to collect the sectors with the maximum visit count.

Thus, the overall complexity is O(n * m), which is efficient given that n and m are typically not extraordinarily large based on problem constraints.

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