Minimum cost for tickets

Time: O(N); Space: O(1); medium

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.

Constraints:

  • 1 <= len(days) <= 365

  • 1 <= days[i] <= 365

  • days is in strictly increasing order.

  • len(costs) == 3

  • 1 <= costs[i] <= 1000

1. Dynamic Programming (Day Variant) [O(W), O(W)]

Intuition and Algorithm

For each day, if you don’t have to travel today, then it’s strictly better to wait to buy a pass. If you have to travel today, you have up to 3 choices: you must buy either a 1-day, 7-day, or 30-day pass.

We can express those choices as a recursion and use dynamic programming. Let’s say dp(i) is the cost to fulfill your travel plan from day i to the end of the plan. Then, if you have to travel today, your cost is:

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

[14]:

from functools import lru_cache

class Solution1(object):
    """
    Time: O(W), where W = 365 is the maximum numbered day in your travel plan
    Space: O(W)
    """
    def mincostTickets(self, days, costs):
        """
        :type days: List[int]
        :type costs: List[int]
        :rtype: int
        """
        dayset = set(days)
        durations = [1, 7, 30]

        @lru_cache(None)
        def dp(i):
            if i > 365:
                return 0
            elif i in dayset:
                return min(dp(i + d) + c
                           for c, d in zip(costs, durations))
            else:
                return dp(i + 1)

        return dp(1)

[15]:

s = Solution1()

days = [1,4,6,7,8,20]
costs = [2,7,15]
assert s.mincostTickets(days, costs) == 11

days = [1,2,3,4,5,6,7,8,9,10,30,31]
costs = [2,7,15]
assert s.mincostTickets(days, costs) == 17

2. Dynamic Programming (Window Variant) [O(N), O(N)]

Intuition and Algorithm

As in Approach 1, we only need to buy a travel pass on a day we intend to travel.

Now, let dp(i) be the cost to travel from day days[i] to the end of the plan. If say, j1 is the largest index such that days[j1] < days[i] + 1, j7 is the largest index such that days[j7] < days[i] + 7, and j30 is the largest index such that days[j30] < days[i] + 30, then we have:

dp(i) = min(dp(j1) + costs}[0], dp(j7) + costs[1], dp(j30) + costs[2]

[16]:

from functools import lru_cache

class Solution2(object):
    """
    Time: O(N), where N is the number of unique days in your travel plan
    Space: O(N)
    """
    def mincostTickets(self, days, costs):
        """
        :type days: List[int]
        :type costs: List[int]
        :rtype: int
        """
        N = len(days)
        durations = [1, 7, 30]

        @lru_cache(None)
        def dp(i):               # How much money to do days[i]+
            if i >= N: return 0

            ans = float('inf')
            j = i
            for c, d in zip(costs, durations):
                while j < N and days[j] < days[i] + d:
                    j += 1
                ans = min(ans, dp(j) + c)

            return ans

        return dp(0)

[17]:

s = Solution2()

days = [1,4,6,7,8,20]
costs = [2,7,15]
assert s.mincostTickets(days, costs) == 11

days = [1,2,3,4,5,6,7,8,9,10,30,31]
costs = [2,7,15]
assert s.mincostTickets(days, costs) == 17

3. Dynamic Programming [O(N), O(1)]

[18]:

class Solution3(object):
    """
    Time: O(N)
    Space: O(1)
    """
    def mincostTickets(self, days, costs):
        """
        :type days: List[int]
        :type costs: List[int]
        :rtype: int
        """
        durations = [1, 7, 30]
        W = durations[-1]
        dp = [float("inf") for i in range(W)]
        dp[0] = 0
        last_buy_days = [0, 0, 0]

        for i in range(1,len(days)+1):
            dp[i%W] = float("inf")
            for j in range(len(durations)):
                while i-1 < len(days) and \
                      days[i-1] > days[last_buy_days[j]]+durations[j]-1:
                    last_buy_days[j] += 1  # Time: O(n)
                dp[i%W] = min(dp[i%W], dp[last_buy_days[j]%W]+costs[j])

        return dp[len(days)%W]

[19]:

s = Solution3()

days = [1,4,6,7,8,20]
costs = [2,7,15]
assert s.mincostTickets(days, costs) == 11

days = [1,2,3,4,5,6,7,8,9,10,30,31]
costs = [2,7,15]
assert s.mincostTickets(days, costs) == 17