## Description

In a country popular for train travel, you have planned some train travelling one year in advance. The days of the year that you will travel is given as an array `days`

. Each day is an integer from `1`

to `365`

.

Train tickets are sold in 3 different ways:

- a 1-day pass is sold for
`costs[0]`

dollars; - a 7-day pass is sold for
`costs[1]`

dollars; - 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: day 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, 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.
```

**Note:**

`1 <= days.length <= 365`

`1 <= days[i] <= 365`

`days`

is in strictly increasing order.`costs.length == 3`

`1 <= costs[i] <= 1000`

## Solutions

### 1. DP

```
# Time: O(n)
# Space: O(n)
class Solution:
def mincostTickets(self, days: List[int], costs: List[int]) -> int:
travel_days = set(days)
dp = [0 for i in range(days[-1] + 1)]
for day_i in range(1, days[-1] + 1):
if day_i not in travel_days:
dp[day_i] = dp[day_i - 1]
continue
dp[day_i] = min(dp[max(0, day_i - 1)] + costs[0],
dp[max(0, day_i - 7)] + costs[1],
dp[max(0, day_i - 30)] + costs[2])
return dp[-1]
# 66/66 cases passed (36 ms)
# Your runtime beats 89.82 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.8 MB)
```