## Description

https://leetcode.com/problems/distance-between-bus-stops/

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 = 1Output:1Explanation:Distance between 0 and 1 is 1 or 9, minimum is 1.

**Example 2:**

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

**Example 3:**

Input:distance = [1,2,3,4], start = 0, destination = 3Output:4Explanation: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`

## Explanation

Two scenarios, one is: start < destination, the other is: start > destination. In both scenarios, we check whether the minimum distance is either clockwise or counterclockwise.

## Python Solution

```
class Solution:
def distanceBetweenBusStops(self, distance: List[int], start: int, destination: int) -> int:
if start < destination:
min_distance = min(sum(distance[start:destination]), (sum(distance[destination:]) + sum(distance[:start])))
else:
min_distance = min(sum(distance[destination:start]), (sum(distance[start:]) + sum(distance[:destination])))
return min_distance
```

- Time Complexity: O(N).
- Space Complexity: O(1).