Swift Coding Test Course, Travel Planning

Planning a trip can be an exciting yet challenging task for many people. Various elements such as destination selection, itinerary, and budget management must be considered. In this course, we will explore how to solve these issues through programming.

Problem Definition

Please create an algorithm to plan a trip that satisfies the following conditions.

• A list of destinations is provided.
• Each destination has a specific cost and time required.
• You want to visit as many destinations as possible without exceeding a maximum travel budget (B) within a fixed duration (C).
• Output the combinations of destinations that can be visited.

Input and Output Format

Input:

    Number of destinations N (1 ≤ N ≤ 20)
    Cost and time required for each destination (cost, time)
    Maximum travel budget B (1 ≤ B ≤ 1000)
    Maximum time limit C (1 ≤ C ≤ 100)
    

Output:

    List of possible maximum destinations
    

Example

    Input:
    4
    100 1
    200 2
    150 1
    300 4
    400
    5

    Output:
    Destinations: (100, 1), (150, 1), (200, 2)
    

Problem-Solving Approach

This problem can be classified as a combination problem since it requires finding possible combinations with given resources (cost, time).

To solve this, we will use the Backtracking technique. Backtracking is a method that explores all possible cases to find a solution. For each destination, we will have two choices:

• Visit the destination.
• Do not visit the destination.

We will check the cost and time at each path to ensure we do not exceed the maximum limits. This process will be repeated recursively to explore all combinations.

Swift Code Implementation

Now let’s implement the above algorithm in Swift.

    struct Destination {
        let cost: Int
        let time: Int
    }
    
    func planTrip(destinations: [Destination], budget: Int, timeLimit: Int) -> [(Int, Int)] {
        var maxDestinations: [(Int, Int)] = []
        
        func backtrack(index: Int, currentBudget: Int, currentTime: Int, currentList: [(Int, Int)]) {
            if currentBudget > budget || currentTime > timeLimit {
                return
            }
            if currentList.count > maxDestinations.count {
                maxDestinations = currentList
            }
            for i in index..