983. Minimum Cost For Tickets

Problem

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.

https://leetcode.com/problems/minimum-cost-for-tickets/

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, 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

Test Cases

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class Solution:
    def mincostTickets(self, days: List[int], costs: List[int]) -> int:

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import pytest

from solution import Solution


@pytest.mark.parametrize('days, costs, expected', [
    ([1,4,6,7,8,20], [2,7,15], 11),
    ([1,2,3,4,5,6,7,8,9,10,30,31], [2,7,15], 17),

    ([1,4,6,7,8,20], [7,2,15], 6),
    ([1,2,3,4,6,8,9,10,13,14,16,17,19,21,24,26,27,28,29], [3,14,50], 50),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, days, costs, expected):
    assert sol.mincostTickets(days, costs.copy()) == expected

Thoughts

如果某一天要出行，要么当天买单日票，要么直接用最早 6 天前买的七日票，要么直接用最早 29 天前买的三十日票。如果不出行，则当天不再额外花钱。

定义 dp(d) 表示截止到第 d（1-indexed）天的最低旅行成本，那么有：

$$dp(d)=\begin{cases} dp(d-1) & \text{if not travel on day d} \\ \min\begin{cases} dp(d-1)+costs[0] \\ dp(d-7)+costs[1] \\ dp(d-30)+costs[2] \end{cases} \end{cases}$$

定义 dp(∀d < 1) = 0。然后从 d = 1 递推到最后一个旅行日（days[-1]），最终结果为 dp(days[-1])。

因为最多只需要用到 30 天前的 dp 值，可以像 2466. Count Ways To Build Good Strings 那样用队列保持最新的至多 30 个值，节省一点儿空间。

时间复杂度 O(days[-1])，空间复杂度 O(1)（30 记为常数）。

Code

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from collections import deque


class Solution:
    def mincostTickets(self, days: list[int], costs: list[int]) -> int:
        max2 = lambda a, b: a if a >= b else b
        last_day = days[-1]
        dp = deque([0], maxlen=30)
        i = 0
        for d in range(1, last_day + 1):
            if days[i] == d:
                i += 1
                dp.append(min(
                    dp[-1] + costs[0],
                    dp[max2(-7, -d)] + costs[1],
                    dp[max2(-30, -d)] + costs[2],
                ))
            else:
                dp.append(dp[-1])

        return dp[-1]

参考资料

• 本文引用

  • 2466. Count Ways To Build Good Strings

许可协议

本文除了题目描述（Problem）部分之外，其他内容由 Calf 原创，采用 CC BY-NC-SA 4.0 许可协议，转载请注明出处。

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