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Distance Between Bus Stops

EasyUpdated: Aug 2, 2025
Practice on:
LeetCode

Problem

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.

Examples

Example 1


![](https://assets.leetcode.com/uploads/2019/09/03/untitled-diagram-1.jpg)

    
    
    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


![](https://assets.leetcode.com/uploads/2019/09/03/untitled-diagram-1-1.jpg)

    
    
    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


![](https://assets.leetcode.com/uploads/2019/09/03/untitled-diagram-1-2.jpg)

    
    
    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

Solution

Method 1 – Prefix Sum and Circular Traversal

Intuition

The shortest distance between two stops on a circular route is the minimum of the clockwise and counterclockwise distances. We can use prefix sums to quickly compute the distance in either direction.

Approach

  1. If start > destination, swap them so start <= destination.
  2. Compute the clockwise distance as the sum of distance[start:destination].
  3. Compute the total distance as the sum of all elements in distance.
  4. The counterclockwise distance is total - clockwise.
  5. Return the minimum of the two distances.

Code

Java
class Solution {
    public int distanceBetweenBusStops(int[] distance, int start, int destination) {
        if (start > destination) {
            int tmp = start; start = destination; destination = tmp;
        }
        int clockwise = 0, total = 0;
        for (int i = 0; i < distance.length; ++i) {
            total += distance[i];
            if (i >= start && i < destination) clockwise += distance[i];
        }
        return Math.min(clockwise, total - clockwise);
    }
}
Python
class Solution:
    def distanceBetweenBusStops(self, distance: list[int], start: int, destination: int) -> int:
        if start > destination:
            start, destination = destination, start
        clockwise = sum(distance[start:destination])
        total = sum(distance)
        return min(clockwise, total - clockwise)

Complexity

  • ⏰ Time complexity: O(n), where n is the number of stops. We sum the distances once.
  • 🧺 Space complexity: O(1), as only a few variables are used.

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