LeetCode #1184 — EASY

Distance Between Bus Stops

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

The Problem

Problem Statement

A bus has n stops numbered from 0 to n - 1 that form a circle. We know the distance between all pairs of neighboring stops where distance[i] is the distance between the stops number i and (i + 1) % n.

The bus goes along both directions i.e. clockwise and counterclockwise.

Return the shortest distance between the given start and destination stops.

Example 1:

Input: distance = [1,2,3,4], start = 0, destination = 1
Output: 1
Explanation: Distance between 0 and 1 is 1 or 9, minimum is 1.

Example 2:

Input: distance = [1,2,3,4], start = 0, destination = 2
Output: 3
Explanation: Distance between 0 and 2 is 3 or 7, minimum is 3.

Example 3:

Input: distance = [1,2,3,4], start = 0, destination = 3
Output: 4
Explanation: Distance between 0 and 3 is 6 or 4, minimum is 4.

Constraints:

1 <= n <= 10^4
distance.length == n
0 <= start, destination < n
0 <= distance[i] <= 10^4

Step 01

Brute Force Baseline

Problem summary: A bus has n stops numbered from 0 to n - 1 that form a circle. We know the distance between all pairs of neighboring stops where distance[i] is the distance between the stops number i and (i + 1) % n. The bus goes along both directions i.e. clockwise and counterclockwise. Return the shortest distance between the given start and destination stops.

Baseline thinking

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

Pattern signal: Array

Example 1

[1,2,3,4]
0
1

Example 2

[1,2,3,4]
0
2

Example 3

[1,2,3,4]
0
3

Core Insight

What unlocks the optimal approach

Find the distance between the two stops if the bus moved in clockwise or counterclockwise directions.

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 #1184: Distance Between Bus Stops
class Solution {
    public int distanceBetweenBusStops(int[] distance, int start, int destination) {
        int s = Arrays.stream(distance).sum();
        int n = distance.length, t = 0;
        while (start != destination) {
            t += distance[start];
            start = (start + 1) % n;
        }
        return Math.min(t, s - t);
    }
}

// Accepted solution for LeetCode #1184: Distance Between Bus Stops
func distanceBetweenBusStops(distance []int, start int, destination int) int {
	s, t := 0, 0
	for _, x := range distance {
		s += x
	}
	for start != destination {
		t += distance[start]
		start = (start + 1) % len(distance)
	}
	return min(t, s-t)
}

# Accepted solution for LeetCode #1184: Distance Between Bus Stops
class Solution:
    def distanceBetweenBusStops(
        self, distance: List[int], start: int, destination: int
    ) -> int:
        s = sum(distance)
        t, n = 0, len(distance)
        while start != destination:
            t += distance[start]
            start = (start + 1) % n
        return min(t, s - t)

// Accepted solution for LeetCode #1184: Distance Between Bus Stops
impl Solution {
    pub fn distance_between_bus_stops(distance: Vec<i32>, start: i32, destination: i32) -> i32 {
        let s: i32 = distance.iter().sum();
        let mut t = 0;
        let n = distance.len();
        let mut start = start as usize;
        let destination = destination as usize;

        while start != destination {
            t += distance[start];
            start = (start + 1) % n;
        }

        t.min(s - t)
    }
}

// Accepted solution for LeetCode #1184: Distance Between Bus Stops
function distanceBetweenBusStops(distance: number[], start: number, destination: number): number {
    const s = distance.reduce((a, b) => a + b, 0);
    const n = distance.length;
    let t = 0;
    while (start !== destination) {
        t += distance[start];
        start = (start + 1) % n;
    }
    return Math.min(t, s - t);
}

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.