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.

Note:

1. 1 <= days.length <= 365
2. 1 <= days[i] <= 365
3. days is in strictly increasing order.
4. costs.length == 3
5. 1 <= costs[i] <= 1000

Solution: DP

dp[i] := min cost to cover the i-th day
dp[0] = 0
dp[i] = min(dp[i – 1] + costs[0], dp[i – 7] + costs[1], dp[i – 30] + costs[2])

// Author: Huahua, running time: 0 ms
class Solution {
public:
  int mincostTickets(vector<int>& days, vector<int>& costs) {
    vector<int> req(days.back() + 1);
    vector<int> dp(days.back() + 1);
    for (int day : days) req[day] = 1;
    dp[0] = 0;
    for (int i = 1; i < dp.size(); ++i) {
      if (!req[i]) {
        dp[i] = dp[i - 1];
        continue;
      }
      dp[i] = dp[i - 1] + costs[0];
      dp[i] = min(dp[i], dp[max(0, i - 7)] + costs[1]);
      dp[i] = min(dp[i], dp[max(0, i - 30)] + costs[2]);
    }
    return dp.back();
  }
};

// Author: Huahua, running time: 60 ms
class Solution:
  def mincostTickets(self, days: 'List[int]', costs: 'List[int]') -> 'int':
    req = set(days)
    dp = [0] * (days[-1] + 1)    
    for i in range(1, len(dp)):
      if not i in req: 
        dp[i] = dp[i - 1]
        continue
      dp[i] = min(dp[max(0, i - 1)] + costs[0],
                  dp[max(0, i - 7)] + costs[1],
                  dp[max(0, i - 30)] + costs[2])
    return dp[-1]

The post 花花酱LeetCode 983. Minimum Cost For Tickets appeared first on Huahua's Tech Road.