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:11Explanation: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:17Explanation: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`

## Solution: DP

dp[i] := min cost to cover the i-th day

dp[0] = 0

dp[i] = min(dp[i – 1] + costs[0], dp[i – 7] + costs[1], dp[i – 30] + costs[2])

## C++

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// Author: Huahua, running time: 0 ms class Solution { public: int mincostTickets(vector<int>& days, vector<int>& costs) { vector<int> req(days.back() + 1); vector<int> dp(days.back() + 1); for (int day : days) req[day] = 1; dp[0] = 0; for (int i = 1; i < dp.size(); ++i) { if (!req[i]) { dp[i] = dp[i - 1]; continue; } dp[i] = dp[i - 1] + costs[0]; dp[i] = min(dp[i], dp[max(0, i - 7)] + costs[1]); dp[i] = min(dp[i], dp[max(0, i - 30)] + costs[2]); } return dp.back(); } }; |

## Python

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// Author: Huahua, running time: 60 ms class Solution: def mincostTickets(self, days: 'List[int]', costs: 'List[int]') -> 'int': req = set(days) dp = [0] * (days[-1] + 1) for i in range(1, len(dp)): if not i in req: dp[i] = dp[i - 1] continue dp[i] = min(dp[max(0, i - 1)] + costs[0], dp[max(0, i - 7)] + costs[1], dp[max(0, i - 30)] + costs[2]) return dp[-1] |

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