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minimum-cost-for-tickets

983. Minimum Cost For Tickets​

You have planned some train traveling one year in advance. The days of the year in which you will travel are given as an integer array days. Each day is an integer from 1 to 365.

Train tickets are sold in three different ways:

  • a 1-day pass is sold for costs[0] dollars,
  • a 7-day pass is sold for costs[1] dollars, and
  • a 30-day pass is sold for costs[2] dollars.

The passes allow that many days of consecutive travel.

  • For example, if we get a 7-day pass on day 2, then we can travel for 7 days: 2, 3, 4, 5, 6, 7, and 8.

Return the minimum number of dollars you need to travel every day in the given list of days.

Example 1:

Input: days = [1,4,6,7,8,20], costs = [2,7,15] Output: 11 Explanation: For example, here is one way to buy passes that lets you travel your travel plan: On day 1, you bought a 1-day pass for costs[0] = 2, which covered day 1. On day 3, you bought a 7-day pass for costs\[1\] = 7, which covered days 3, 4, ..., 9. On day 20, you bought a 1-day pass for costs[0] = 2,whichcoveredday20.Intotal,youspent2, which covered day 20. In total, you spent 11 and covered all the days of your travel.

Example 2:

Input: days = [1,2,3,4,5,6,7,8,9,10,30,31], costs = [2,7,15] Output: 17 Explanation: For example, here is one way to buy passes that lets you travel your travel plan: On day 1, you bought a 30-day pass for costs[2] = 15 which covered days 1, 2, ..., 30. On day 31, you bought a 1-day pass for costs\[0\] = 2 which covered day 31. In total, you spent $17 and covered all the days of your travel.

Constraints:

  • 1 <= days.length <= 365
  • 1 <= days[i] <= 365
  • days is in strictly increasing order.
  • costs.length == 3
  • 1 <= costs[i] <= 1000
class Solution {
public:
    int mincostTickets(vector<int>& days, vector<int>& costs) {
        int lastDay = days.back();
        vector<int> dp(lastDay + 1, 0);

        // 給 days 數列的指針
        int i = 0;

        // 使用一個 loop 從第一天到旅行的最後一天
        for (int day = 1; day < lastDay + 1; day++) {
            // 如果不是出去玩的那一天
            if (day < days[i]) {
                // 雖然不用花錢，但還是要延續前一天的 DP 值
                dp[day] = dp[day - 1];
            } else {
                // 是要出去玩的那一天耶，喔耶！
                // 順便買票囉！

                // 指針到下一天
                i++;

                // 三種組合求最小的花費
                // 一天前買日票
                // 一週前買週票
                // 30 天前買月票
                dp[day] = min({
                    dp[day - 1] + costs[0],
                    dp[max(0, day - 7)] + costs[1],
                    dp[max(0, day - 30)] + costs[2]});
            }
        }
        return dp[lastDay];
    }
};
  • T: O(N)O(N)
  • S: O(N)O(N)