1. The Problem
You have `T` minutes before an amusement park with `n` rides closes. For each ride `i`, you know its duration `r_i` and the waiting time `W(i, t)` if you get in line at time `t` (where `t` is minutes before closing). The goal is to design a dynamic programming algorithm to find the maximum number of rides you can complete.
2. The Idea: Dynamic Programming
The key to this problem is to define the state based on the time remaining. Since the waiting time for a ride depends on when you arrive, the subproblems must be structured around time.
Let `dp[t]` be the maximum number of rides that can be completed given that there are exactly `t` minutes remaining until the park closes. Our goal is to find `dp[T]`.
- Base Case: With 0 minutes left, you can complete 0 rides. So, `dp[0] = 0`.
- Recurrence Relation: To compute `dp[t]`, we consider the first ride we could take at this moment. For any ride `i`, the total time it would consume is its waiting time at time `t`, `W(i, t)`, plus its duration, `r_i`. Let `time_cost = W(i, t) + r_i`. If we take ride `i`, we use up `time_cost` minutes, and we are left with `t - time_cost` minutes. The maximum number of rides we can take after that is given by our subproblem solution, `dp[t - time_cost]`. Therefore, the total number of rides if we choose `i` first is `1 + dp[t - time_cost]`. We should try this for every ride `i` and take the maximum.
This maximum is taken over all rides `i` such that `W(i, t) + r_i <= t`. If no ride can be taken, `dp[t]` is 0 (or simply the value from `dp[t-1]`, implying we wait a minute).
3. The Algorithm
- Create a DP array `dp` of size `T+1`, and initialize all its entries to 0.
- Iterate `t` from 1 to `T` (for each minute before closing).
- For each `t`, initialize its value with the previous minute's result: `dp[t] = dp[t-1]`. This represents the choice of not starting a ride at this exact minute.
- Inner loop for `i` from 1 to `n` (for each ride).
- Calculate the total time cost for ride `i` if started at time `t`: `time_cost = W(i, t) + r_i`.
- If `time_cost <= t`:
- Calculate the potential number of rides: `num_rides = 1 + dp[t - time_cost]`.
- Update `dp[t] = max(dp[t], num_rides)`.
- The maximum number of rides is `dp[T]`.
4. Complexity Analysis
- Time Complexity: O(n * T)
- The algorithm has two nested loops. The outer loop for time `t` runs `T` times. The inner loop for rides `i` runs `n` times. The operations inside are constant time (assuming `W(i, t)` is a table lookup). This gives a total time complexity of O(n * T).
- Space Complexity: O(T)
- We use a DP array `dp` of size `T+1` to store the solutions to subproblems, which requires O(T) space.