Minimum Cost For Tickets

class Solution {
public:
    int travelCostOptimizer(vector<int>& travelDays, vector<int>& ticketCosts) {
        int maxTravelDay = travelDays.back();
        vector<int> minCost(maxTravelDay + 1, 0);
        
        int travelIndex = 0;
        for (int currentDay = 1; currentDay <= maxTravelDay; currentDay++) {
            if (currentDay < travelDays[travelIndex]) {
                minCost[currentDay] = minCost[currentDay - 1];
            } else {
                travelIndex++;
                minCost[currentDay] = min({minCost[currentDay - 1] + ticketCosts[0],
                                           minCost[max(0, currentDay - 7)] + ticketCosts[1],
                                           minCost[max(0, currentDay - 30)] + ticketCosts[2]});
            }
        }
     
        return minCost[maxTravelDay];
    }
};

This solution focuses on determining the minimum cost to travel across a set number of days using a dynamic programming method. The function travelCostOptimizer takes two vectors as input: travelDays, which contains the specific days on which travel occurs, and ticketCosts, which represents the cost of different types of tickets that one can purchase. These ticket types cover single day, seven days, and thirty days of travel, respectively.

The approach involves creating a vector minCost where each index represents the minimum cost to travel up to that day. The algorithm iterates through each day, using previous computed values to determine the least cost for a given day either by using a single day pass, a seven day pass or a thirty day pass. If a particular day is not a travel day, then the cost remains the same as the previous day.

• Steps to determine the minimum cost:
  1. Define maxTravelDay to find the last day of travel from travelDays.
  2. Initialize a vector minCost to store minimum costs up to each day, starting from zero up to maxTravelDay.
  3. Use a loop to fill minCost where for each day, consider if it is a travel day or not.
  4. If it is a travel day, compute the minimum cost by considering the cheapest option between buying a single day, seven days, or thirty days ticket.
  5. Return the value at maxTravelDay in minCost as the minimum cost required for the traveling days noted.

The function results in an efficient computation of the minimum traveling cost, leveraging past computed costs in a time-efficient manner using dynamic programming.

class Solution {
    public int minimumCostTravel(int[] travelDays, int[] passCosts) {
        // Identify the furthest day required for travel.
        int finalDay = travelDays[travelDays.length - 1];
        int travelCosts[] = new int[finalDay + 1];
        Arrays.fill(travelCosts, 0);

        int dayIndex = 0;
        for (int currentDay = 1; currentDay <= finalDay; currentDay++) {
            // Ensure costs are maintained on non-travel days.
            if (currentDay < travelDays[dayIndex]) {
                travelCosts[currentDay] = travelCosts[currentDay - 1];
            } else {
                // Progress to the next required travel day after ticket purchase.
                dayIndex++;
                // Calculate the most cost-effective ticket purchase.
                travelCosts[currentDay] = Math.min(travelCosts[currentDay - 1] + passCosts[0],
                        Math.min(travelCosts[Math.max(0, currentDay - 7)] + passCosts[1],
                                travelCosts[Math.max(0, currentDay - 30)] + passCosts[2]));
            }
        }

        return travelCosts[finalDay];
    }
}

For the problem titled "Minimum Cost For Tickets," the solution calculates the least expensive way to purchase travel passes based on specific travel days and the costs of three types of passes: 1-day, 7-day, and 30-day. The solution is implemented in Java.

Here's a breakdown of the solution approach:

• Start by finding the last day in your input travelDays array, denoted as finalDay. This lets you know until which day you need to consider travel costs.
• Create an array travelCosts sized to finalDay + 1, initializing all elements to zero. This array keeps track of the minimum travel costs accrued by each day up to finalDay.
• Iterate over each day from 1 to finalDay. For days not in the travelDays array (non-travel days), copy the previous day's cost as no new pass is required.
• On days you need to travel, decide which pass to buy by comparing the costs of continuing with a pass bought on an earlier day versus buying a new pass. This involves comparing costs as follows:
  • A new 1-day pass from the previous day's total cost plus the cost of the 1-day pass.
  • A new 7-day pass from the cost 7 days ago plus the cost of the 7-day pass; if less than 7 days have elapsed, compare from day 0.
  • A new 30-day pass from the cost 30 days ago plus the cost of the 30-day pass; if less than 30 days have elapsed, compare from day 0.
• Store the minimum result of these compares in the travelCosts array.

By the end of the loop, travelCosts[finalDay] contains the minimum cost needed to cover all travel days efficiently. The dynamic programming approach ensures that each day is handled in an optimal way based on previous decisions, leading to an overall minimal travel expense.