The “Most Visited Sector in a Circular Track” problem is a classic problem featured on Leetcode that falls under the category of array manipulation and basic algorithmic design. This problem not only assesses your understanding of arrays but also evaluates your ability to think in terms of circular data structures. This article provides a comprehensive guide to tackling this problem, offering multiple solutions and analyzing them for time and space complexity.

## Problem Statement

The problem asks us to determine the most visited sectors in a circular race track consisting of `n`

sectors numbered from 1 to `n`

. A runner starts at sector 1 and runs the track, visiting sectors in the given order. The task is to return the sectors that are most visited in ascending order.

### Constraints:

`2 <= n <= 100`

`1 <= rounds.length <= 100`

`1 <= rounds[i] <= n`

`rounds[i] != rounds[i + 1]`

for`0 <= i < rounds.length - 1`

### Example

For `n = 4`

, and `rounds = [1, 3, 1, 2]`

, the most visited sectors would be `[1, 2]`

.

## Approach 1: Brute-Force Counting

### Algorithm

- Create an array
`count`

of size`n+1`

to keep track of the number of visits for each sector. - Traverse the
`rounds`

array, updating the count for each sector visited. - Find the maximum count and filter out sectors with that count.

#### Python Implementation

```
def mostVisited(n, rounds):
count = [0] * (n + 1)
# Initialize the first sector
count[rounds[0]] = 1
cur = rounds[0]
# Count each sector visited
for i in range(1, len(rounds)):
while cur != rounds[i]:
cur = cur + 1 if cur < n else 1
count[cur] += 1
# Find the maximum count
max_count = max(count)
# Return sectors with the maximum count
return [i for i, c in enumerate(count) if c == max_count]
```

### Time Complexity

The time complexity is O(n×m), where n is the number of sectors and m is the length of the `rounds`

array.

### Space Complexity

The space complexity is O(n).

## Approach 2: Optimized Counting

### Algorithm

- Realize that visiting sector
`i`

to sector`j`

consecutively means all the sectors in between are also visited once. - Use this observation to update the count array more efficiently.

#### Python Implementation

```
def mostVisited(n, rounds):
count = [0] * (n + 1)
# Initialize the first sector
cur = rounds[0]
count[cur] = 1
# Count each sector visited in an optimized way
for i in range(1, len(rounds)):
next_sector = rounds[i]
while cur != next_sector:
cur = cur % n + 1 # Go to the next sector in a circular manner
count[cur] += 1
# Handle the last sector separately
count[rounds[-1]] += 1
# Find the maximum count
max_count = max(count)
# Return sectors with the maximum count
return [i for i, c in enumerate(count) if c == max_count]
```

### Time Complexity

The time complexity is O(m).

### Space Complexity

The space complexity is O(n).

## Comparison of Approaches

## Conclusion

The “Most Visited Sector in a Circular Track” problem serves as a great exercise for practicing array manipulations and understanding circular data structures. Both approaches discussed in this article have their own merits. The brute-force method is easier to implement but comes at the cost of efficiency. On the other hand, the optimized approach is fast but requires a deeper understanding of the problem.