[LeetCode] 983. Minimum Cost For Tickets (Python)

 

난이도: Medium

 

문제 설명

---

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.

 

문제 예제

---

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.

 

제한사항

---

• 1 <= days.length <= 365
• 1 <= days[i] <= 365
• days is in strictly increasing order.
• costs.length == 3
• 1 <= costs[i] <= 1000

 

✏️Solution(솔루션)

class Solution:
    def mincostTickets(self, days: List[int], costs: List[int]) -> int:
        dp = [0 for _ in range(days[-1]+1)]
        
        day_check = defaultdict(bool)
        for day in days:
            day_check[day] = True
        

        for i in range(1, days[-1]+1):
            if not day_check[i]:
                dp[i] = dp[i-1]
            else:
                if i-7 < 0:
                    dp[i] = min(dp[i-1] + costs[0], costs[1], costs[2])
                elif i-30 < 0:
                    dp[i] = min(dp[i-1] + costs[0], dp[i-7] + costs[1], costs[2])
                else:
                    dp[i] = min(dp[i-1] + costs[0], dp[i-7] + costs[1], dp[i-30] + costs[2])

        return dp[-1]

 

이 문제 역시 DP로 해결했다.

 

days에 있는 마지막 day까지 길이의 dp 리스트를 만들었다.

 

그리고 1부터 마지막 day까지 for문을 돌면서 이전날 최소 비용(dp[i-1])에 cost[0]을 더한 값, 7일전 최소 비용(dp[i-7])에 cost[1]을 더한 값, 30일전 최소 비용(dp[i-30])에 cost[2]를 더한 값의 최소값을 찾아 현재날에 최소 비용(dp[i])을 업데이트했다.

 

마지막에는 dp에 저장된 마지막 요소를 return 하여 정답을 맞출 수 있었다.

---

문제: 983. Minimum Cost For Tickets

 

깃허브: github

algorithm_practice/LeetCode/983-minimum-cost-for-tickets at main · laewonJeong/algorithm_practice