# 983. Minimum Cost For Tickets
###### tags: `leetcode`
## Description
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, 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:

>$1 \leq days.length \leq 365$
$1 \leq days[i] \leq 365$
days is in strictly increasing order.
costs.length == 3
$1 \leq costs[i] \leq 1000$

## Solution
- The problem can be solved by using dynamic programming
- To keep the record from backward, meaning that `dp[i]` is the minimum combination from the last day to the `days[i]`. The answer would be `dp[0]`

dp.resize(days.size());
minDFS(days, costs, 0);
return dp[0];

- The iteration for constructing the array can be done by choosing the three options and check the lower bound date after using the ticket. Finally compare the value to get the smallest one

if (dp[idx] == 0)
{
    // 1 day ticket
    if (idx + 1 != days.size())
    {
        minDFS(days, costs, idx + 1);
        dp[idx] = dp[idx + 1] + costs[0];
    }
    else dp[idx] = costs[0];
    // 7 day ticket
    auto it = lower_bound(days.begin() + idx, days.end(), days[idx] + 7);
    if (it != days.end())
    {
        minDFS(days, costs, it - days.begin());
        dp[idx] = min(dp[idx], dp[it - days.begin()] + costs[1]);
    }
    else dp[idx] = min(dp[idx], costs[1]);
    // 30 day ticket
    it = lower_bound(days.begin() + idx, days.end(), days[idx] + 30);
    if (it != days.end())
    {
        minDFS(days, costs, it - days.begin());
        dp[idx] = min(dp[idx], dp[it - days.begin()] + costs[2]);
    }
    else dp[idx] = min(dp[idx], costs[2]);            
}