# 983. 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`.

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`

``````class Solution {
public:
int mincostTickets(vector<int>& days, vector<int>& costs) {
int n = days.size();
vector<int> dp(n + 1, INT_MAX);
dp[0] = 0;
for (int i = 1; i <= n; ++i) {
dp[i] = min(dp[i], dp[i - 1] + costs[0]);
for (int j = 1; j <= i; ++j) {
if (days[j - 1] + 7 > days[i - 1]) {
dp[i] = min(dp[i], dp[j - 1] + costs[1]);
}
if (days[j - 1] + 30 > days[i - 1]) {
dp[i] = min(dp[i], dp[j - 1] + costs[2]);
}
}
}
return dp.back();
}
};
``````

``````class Solution {
public:
int mincostTickets(vector<int>& days, vector<int>& costs) {
unordered_set<int> st(days.begin(), days.end());
vector<int> dp(366);
for (int i = 1; i <= 365; ++i) {
dp[i] = dp[i - 1];
if (st.count(i)) {
dp[i] = min({dp[i - 1] + costs[0], dp[max(0, i - 7)] + costs[1], dp[max(0, i - 30)] + costs[2]});
}
}
return dp.back();
}
};
``````

Github 同步地址:

https://github.com/grandyang/leetcode/issues/983

Coin Change

https://leetcode.com/problems/minimum-cost-for-tickets/

https://leetcode.com/problems/minimum-cost-for-tickets/discuss/226659/Two-DP-solutions-with-pictures

https://leetcode.com/problems/minimum-cost-for-tickets/discuss/630868/explanation-from-someone-who-took-2-hours-to-solve

LeetCode All in One 题目讲解汇总(持续更新中…)

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