Minimum Cost For Tickets
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.
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.Let's see an example of this. Given the array:
days = [1, 4, 6, 7, 8, 20]And the ticket costs:
costs = [2, 7, 15]We want to find out the minimum cost to travel everyday in the given list. Essentially, saving us money.
Costs is:
cost[0] = 1 day
cost[1] = 7 days
costs[2] = 30 daysLet's try to build this into a recurrence.
Base case
1 day = cost[0]
Divide & Conquer
Everyday after this is either the minimum cost of the previous days, or the current ticket.
So if we are at 7 days, our cost will either be:
Day 6's ticket (assuming it covers that day)
Day 7's cost[1] ticket
While calculating the min, we need to make sure our ticket does not exceed the days. We can visualise this with a table:
This methodology quickly falls apart, as our recurrence will look like:
Previous days ticket (assuming the day is covered)
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