JavaScript Coding Test Course, Planning a Trip

Hello! In this post, we will discuss a problem that involves planning a trip using JavaScript. This may be helpful for those studying algorithms for employment. Through this problem, you will gain an understanding of array manipulation and search algorithms in JavaScript.

Problem Description

In the process of planning a trip, we want to visit specific cities. The goal of the algorithm problem is to create the most cost-effective travel route based on the distances between the given cities and the final destination.

Problem Definition

Write a function that takes the following input:

    function planTrip(routes, startPoint, endPoint) {
        // Write code here
    }

Input

  • routes: An object containing the distance information between cities
  • startPoint: The starting city of the trip
  • endPoint: The ending city of the trip

Output

Return an object that includes the shortest path and distance to the final city, along with the path.

Example

    const routes = {
        "A": { "B": 5, "C": 10 },
        "B": { "C": 3, "D": 15 },
        "C": { "D": 2 },
        "D": {}
    };
    
    console.log(planTrip(routes, "A", "D"));
    // Output: { distance: 8, path: ["A", "B", "C", "D"] }

How to Solve the Problem

To solve this problem, you should follow these steps:

  1. Store and explore a list of cities that can be visited along with their distances.
  2. Use DFS (Depth-First Search) or BFS (Breadth-First Search) to explore possible travel routes.
  3. Store all paths that reach the final destination and find the shortest distance.
  4. Return the shortest path along with its distance.

Code Implementation

Now let’s write the code according to the above steps:

    function planTrip(routes, startPoint, endPoint) {
        const visited = {};
        const stack = [{ city: startPoint, path: [startPoint], distance: 0 }];
        let shortestPath = null;
        let shortestDistance = Infinity;

        while (stack.length) {
            const { city, path, distance } = stack.pop();

            if (city === endPoint) {
                if (distance < shortestDistance) {
                    shortestDistance = distance;
                    shortestPath = path;
                }
            }
            visited[city] = true;

            for (const neighbor in routes[city]) {
                if (!visited[neighbor]) {
                    stack.push({
                        city: neighbor,
                        path: [...path, neighbor],
                        distance: distance + routes[city][neighbor],
                    });
                }
            }
        }

        return {
            distance: shortestDistance,
            path: shortestPath,
        };
    }

Code Explanation

The code above implements DFS using a stack. It goes through the following steps:

  • Initialization: Initialize variables to track the starting city, current path, distance, and visitation status.
  • Exploration: Pop elements from the stack one by one and add possible routes.
  • Check for Destination Arrival: If the destination is reached, compare the current distance and path to update the shortest distance and path.
  • Return Results: Return the results that include the shortest distance and path.

Complexity Analysis

The time complexity of this algorithm is O(V + E), where V is the number of cities and E is the number of roads. This is because each city and road is explored at least once. The space complexity is also O(V) due to the space used to store visited cities.

Conclusion

Through this problem, I hope you have understood the basics of graph traversal using JavaScript. Planning a trip can be said to be everything about traveling! Use appropriate algorithms to create efficient plans. It is also a good idea to extend this algorithm to solve more complex route optimization problems.

In the next post, we will explore more diverse algorithm problems. Thank you!