#1184

Distance Between Bus Stops

Easy
ArrayArrayPrefix Sum
LeetCode ↗

Approaches

Brute ForceOptimal
Complexity Comparison
Brute ForceOptimal Solution
Time
O(n²)
O(n)
Space
O(1)
O(1)
💡

Intuition

Time O(n)Space O(1)

The optimal solution calculates the distances in a single pass, making it more efficient. By leveraging the circular nature of the bus stops, we can compute both distances directly.

⚙️

Algorithm

5 steps
  1. 1Step 1: Ensure start is less than destination for easier calculations.
  2. 2Step 2: Calculate the clockwise distance by summing the distances from start to destination.
  3. 3Step 3: Calculate the total distance of the entire route.
  4. 4Step 4: Calculate the counterclockwise distance as the total distance minus the clockwise distance.
  5. 5Step 5: Return the minimum of the clockwise and counterclockwise distances.
solution.py9 lines
1# Full working Python code
2
3def distance_between_bus_stops(distance, start, destination):
4    if start > destination:
5        start, destination = destination, start
6    clockwise_distance = sum(distance[start:destination])
7    total_distance = sum(distance)
8    counterclockwise_distance = total_distance - clockwise_distance
9    return min(clockwise_distance, counterclockwise_distance)

Complexity note: The time complexity is O(n) because we only traverse the distance array a couple of times, making it linear with respect to the number of stops.

  • 1Understanding the circular nature of the bus stops is crucial for calculating distances efficiently.
  • 2Recognizing that both clockwise and counterclockwise paths need to be considered helps in finding the shortest distance.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.