## Description

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

## Solutions

### 1. Two Pass

# Time: O(n)
# Space: O(1)
class Solution:
def distanceBetweenBusStops(self, distance: List[int], start: int, destination: int) -> int:
i = start
path1, path2 = 0, 0
n = len(distance)
while i % n != destination:
path1 += distance[i % n]
i += 1
i = start
while i != destination:
i -= 1
if i < 0:
i = n - 1
path2 += distance[i]
return min(path1, path2)

# 37/37 cases passed (40 ms)
# Your runtime beats 91.41 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (13.7 MB)


### 2. One Pass

# Time: O(n)
# Space: O(1)
class Solution:
def distanceBetweenBusStops(self, distance: List[int], start: int, destination: int) -> int:
if start > destination:
start, destination = destination, start

total_sum = 0
dist = 0
n = len(distance)
for i in range(n):
if start <= i < destination:
dist += distance[i]
total_sum += distance[i]
return min(dist, total_sum - dist)

# 37/37 cases passed (44 ms)
# Your runtime beats 75 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (13.7 MB)