0372MediumArrayDynamic Programming

Cheapest Ferry Passes

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

You already know which days of the year you will ride the harbour ferry. trips lists those day numbers in strictly increasing order, counting day 1 as the first day of the year.

The ferry sells three kinds of pass and prices gives their costs in this order:

prices[0] buys a pass valid for 1 day.
prices[1] buys a pass valid for 7 days in a row.
prices[2] buys a pass valid for 30 days in a row.

A pass starts on any day you name and stays valid for its whole span of consecutive days, whether or not you ride on each of them. You may buy as many passes as you like, and every day listed in trips has to fall inside the span of at least one pass.

Return the smallest total you can spend.

Examples

Example 1

Input: trips = [3], prices = [5,20,60]
Output: 5

A 1-day pass on day 3 covers the only ride and costs 5.

Example 2

Input: trips = [5,6,7,8,9,10,11], prices = [4,10,50]
Output: 10

One 7-day pass starting on day 5 stays valid through day 11, so it covers all seven rides, for 10.

Example 3

Input: trips = [12,13,16,21,24,31,37], prices = [50,10,27]
Output: 27

One 30-day pass starting on day 12 stays valid through day 41, which covers every ride, for 27.

Example 4

Input: trips = [15,29,44,52,54,55], prices = [11,6,22]
Output: 24

Four 7-day passes starting on days 15, 29, 44 and 52 cover the rides, and the last of them covers days 52, 54 and 55 together. Four passes at 6 each is 24.

Example 5

Input: trips = [1,10,25,40], prices = [3,10,30]
Output: 12

Four 1-day passes, one on each riding day, cost 12 in total.

Constraints

1 <= trips.length <= 365
1 <= trips[i] <= 365
trips is in strictly increasing order.
prices.length == 3
1 <= prices[i] <= 1000

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def mincost_tickets(trips: list[int], prices: list[int]) -> int:

Java

public int mincostTickets(int[] trips, int[] prices)

The intended approach

Time O(n log n)Space O(n) for the table

Fill a table backwards over the riding days: entry i is the cheapest way to cover trips[i] and everything after it. Whichever pass covers trips[i] can be slid so that it begins on that very day, so for each of the three spans a binary search finds the first riding day the pass fails to reach, and the cost is that pass plus the already-computed entry for that day. The answer is the entry for the first riding day.

Pythonentry point: mincost_tickets

from bisect import bisect_right


def mincost_tickets(trips, prices):
    # Work backwards over the trip list. best[i] is the cheapest way to cover
    # trips[i:], and the pass bought for trips[i] may as well start on that day.
    # Time O(n log n) for the n binary searches, space O(n) for the table.
    spans = (1, 7, 30)
    n = len(trips)
    best = [0] * (n + 1)
    for i in range(n - 1, -1, -1):
        cheapest = None
        for k in range(3):
            last_day = trips[i] + spans[k] - 1
            j = bisect_right(trips, last_day)
            cost = prices[k] + best[j]
            if cheapest is None or cost < cheapest:
                cheapest = cost
        best[i] = cheapest
    return best[0]

The obvious version

Fill a table over every calendar day up to the last ride. A day with no ride costs whatever the previous day cost; a riding day takes the cheapest of three options, each being one pass plus the table entry from the day just before that pass would have started.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_mincost_tickets(trips, prices):
    horizon = trips[-1]
    riding = set(trips)
    table = [0] * (horizon + 1)
    for day in range(1, horizon + 1):
        if day not in riding:
            table[day] = table[day - 1]
            continue
        one = table[day - 1] + prices[0]
        week = table[max(0, day - 7)] + prices[1]
        month = table[max(0, day - 30)] + prices[2]
        table[day] = min(one, week, month)
    return table[horizon]

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
5 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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