LeetCode #1560 — EASY

Most Visited Sector in a Circular Track

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

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 i^th 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

Step 01

Brute Force Baseline

Problem summary: 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).

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

4
[1,3,1,2]

Example 2

2
[2,1,2,1,2,1,2,1,2]

Example 3

7
[1,3,5,7]

Step 02

Core Insight

What unlocks the optimal approach

For each round increment the visits of the sectors visited during the marathon with 1.
Determine the max number of visits, and return any sector visited the max number of visits.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1560: Most Visited Sector in  a Circular Track
class Solution {
    public List<Integer> mostVisited(int n, int[] rounds) {
        int m = rounds.length - 1;
        List<Integer> ans = new ArrayList<>();
        if (rounds[0] <= rounds[m]) {
            for (int i = rounds[0]; i <= rounds[m]; ++i) {
                ans.add(i);
            }
        } else {
            for (int i = 1; i <= rounds[m]; ++i) {
                ans.add(i);
            }
            for (int i = rounds[0]; i <= n; ++i) {
                ans.add(i);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1560: Most Visited Sector in  a Circular Track
func mostVisited(n int, rounds []int) []int {
	m := len(rounds) - 1
	var ans []int
	if rounds[0] <= rounds[m] {
		for i := rounds[0]; i <= rounds[m]; i++ {
			ans = append(ans, i)
		}
	} else {
		for i := 1; i <= rounds[m]; i++ {
			ans = append(ans, i)
		}
		for i := rounds[0]; i <= n; i++ {
			ans = append(ans, i)
		}
	}
	return ans
}

# Accepted solution for LeetCode #1560: Most Visited Sector in  a Circular Track
class Solution:
    def mostVisited(self, n: int, rounds: List[int]) -> List[int]:
        if rounds[0] <= rounds[-1]:
            return list(range(rounds[0], rounds[-1] + 1))
        return list(range(1, rounds[-1] + 1)) + list(range(rounds[0], n + 1))

// Accepted solution for LeetCode #1560: Most Visited Sector in  a Circular Track
struct Solution;

impl Solution {
    fn most_visited(n: i32, rounds: Vec<i32>) -> Vec<i32> {
        let m = rounds.len();
        let start = rounds[0];
        let end = rounds[m - 1];
        if start == end {
            return vec![start];
        }
        if start < end {
            return (start..=end).collect();
        }
        let mut res: Vec<i32> = (start..=end + n).map(|x| ((x - 1) % n) + 1).collect();
        res.sort_unstable();
        res
    }
}

#[test]
fn test() {
    let n = 4;
    let rounds = vec![1, 3, 1, 2];
    let res = vec![1, 2];
    assert_eq!(Solution::most_visited(n, rounds), res);
    let n = 2;
    let rounds = vec![2, 1, 2, 1, 2, 1, 2, 1, 2];
    let res = vec![2];
    assert_eq!(Solution::most_visited(n, rounds), res);
    let n = 7;
    let rounds = vec![1, 3, 5, 7];
    let res = vec![1, 2, 3, 4, 5, 6, 7];
    assert_eq!(Solution::most_visited(n, rounds), res);
    let n = 3;
    let rounds = vec![3, 2, 1, 2, 1, 3, 2, 1, 2, 1, 3, 2, 3, 1];
    let res = vec![1, 3];
    assert_eq!(Solution::most_visited(n, rounds), res);
}

// Accepted solution for LeetCode #1560: Most Visited Sector in  a Circular Track
function mostVisited(n: number, rounds: number[]): number[] {
    const ans: number[] = [];
    const m = rounds.length - 1;
    if (rounds[0] <= rounds[m]) {
        for (let i = rounds[0]; i <= rounds[m]; ++i) {
            ans.push(i);
        }
    } else {
        for (let i = 1; i <= rounds[m]; ++i) {
            ans.push(i);
        }
        for (let i = rounds[0]; i <= n; ++i) {
            ans.push(i);
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.