983. Minimum Cost For Tickets
Description
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 for7days:2,3,4,5,6,7, and8.
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 <= days.length <= 3651 <= days[i] <= 365daysis in strictly increasing order.costs.length == 31 <= costs[i] <= 1000
Solutions
Solution 1: Memoization Search + Binary Search
We define a function $\textit{dfs(i)}$, which represents the minimum cost required from the $i$-th trip to the last trip. Thus, the answer is $\textit{dfs(0)}$.
The execution process of the function $\textit{dfs(i)}$ is as follows:
If $i \geq n$, it means all trips have ended, return $0$;
Otherwise, we need to consider three types of purchases: buying a 1-day pass, buying a 7-day pass, and buying a 30-day pass. We calculate the cost for these three purchasing methods separately and use binary search to find the index $j$ of the next trip, then recursively call $\textit{dfs(j)}$, and finally return the minimum cost among these three purchasing methods.
To avoid repeated calculations, we use memoization search to save the results that have already been calculated.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ represents the number of trips.
Python3
class Solution:
def mincostTickets(self, days: List[int], costs: List[int]) -> int:
@cache
def dfs(i: int) -> int:
if i >= n:
return 0
ans = inf
for c, v in zip(costs, valid):
j = bisect_left(days, days[i] + v)
ans = min(ans, c + dfs(j))
return ans
n = len(days)
valid = [1, 7, 30]
return dfs(0)
Java
class Solution {
private final int[] valid = {1, 7, 30};
private int[] days;
private int[] costs;
private Integer[] f;
private int n;
public int mincostTickets(int[] days, int[] costs) {
n = days.length;
f = new Integer[n];
this.days = days;
this.costs = costs;
return dfs(0);
}
private int dfs(int i) {
if (i >= n) {
return 0;
}
if (f[i] != null) {
return f[i];
}
f[i] = Integer.MAX_VALUE;
for (int k = 0; k < 3; ++k) {
int j = Arrays.binarySearch(days, days[i] + valid[k]);
j = j < 0 ? -j - 1 : j;
f[i] = Math.min(f[i], dfs(j) + costs[k]);
}
return f[i];
}
}
C++
class Solution {
public:
int mincostTickets(vector<int>& days, vector<int>& costs) {
int valid[3] = {1, 7, 30};
int n = days.size();
int f[n];
memset(f, 0, sizeof(f));
function<int(int)> dfs = [&](int i) {
if (i >= n) {
return 0;
}
if (f[i]) {
return f[i];
}
f[i] = INT_MAX;
for (int k = 0; k < 3; ++k) {
int j = lower_bound(days.begin(), days.end(), days[i] + valid[k]) - days.begin();
f[i] = min(f[i], dfs(j) + costs[k]);
}
return f[i];
};
return dfs(0);
}
};
Go
func mincostTickets(days []int, costs []int) int {
valid := [3]int{1, 7, 30}
n := len(days)
f := make([]int, n)
var dfs func(int) int
dfs = func(i int) int {
if i >= n {
return 0
}
if f[i] > 0 {
return f[i]
}
f[i] = 1 << 30
for k := 0; k < 3; k++ {
j := sort.SearchInts(days, days[i]+valid[k])
f[i] = min(f[i], dfs(j)+costs[k])
}
return f[i]
}
return dfs(0)
}
TypeScript
function mincostTickets(days: number[], costs: number[]): number {
const n = days.length;
const f: number[] = Array(n).fill(0);
const valid: number[] = [1, 7, 30];
const search = (x: number): number => {
let [l, r] = [0, n];
while (l < r) {
const mid = (l + r) >> 1;
if (days[mid] >= x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
};
const dfs = (i: number): number => {
if (i >= n) {
return 0;
}
if (f[i]) {
return f[i];
}
f[i] = Infinity;
for (let k = 0; k < 3; ++k) {
const j = search(days[i] + valid[k]);
f[i] = Math.min(f[i], dfs(j) + costs[k]);
}
return f[i];
};
return dfs(0);
}
Solution 2: Dynamic Programming
Let's denote the last day in the $\textit{days}$ array as $m$. We can define an array $f$ of length $m + 1$, where $f[i]$ represents the minimum cost from day $1$ to day $i$.
We can calculate the value of $f[i]$ in increasing order of the dates in the $\textit{days}$ array, starting from day $1$. If day $i$ is a travel day, we can consider three purchasing options: buying a 1-day pass, buying a 7-day pass, and buying a 30-day pass. We calculate the cost for these three purchasing methods separately and take the minimum cost among these three as the value of $f[i]$. If day $i$ is not a travel day, then $f[i] = f[i - 1]$.
The final answer is $f[m]$.
The time complexity is $O(m)$, and the space complexity is $O(m)$. Here, $m$ represents the last day of travel.
Python3
class Solution:
def mincostTickets(self, days: List[int], costs: List[int]) -> int:
m = days[-1]
f = [0] * (m + 1)
valid = [1, 7, 30]
j = 0
for i in range(1, m + 1):
if i == days[j]:
f[i] = inf
for c, v in zip(costs, valid):
f[i] = min(f[i], f[max(0, i - v)] + c)
j += 1
else:
f[i] = f[i - 1]
return f[m]
Java
class Solution {
public int mincostTickets(int[] days, int[] costs) {
int m = days[days.length - 1];
int[] f = new int[m + 1];
final int[] valid = {1, 7, 30};
for (int i = 1, j = 0; i <= m; ++i) {
if (i == days[j]) {
f[i] = Integer.MAX_VALUE;
for (int k = 0; k < 3; ++k) {
int c = costs[k], v = valid[k];
f[i] = Math.min(f[i], f[Math.max(0, i - v)] + c);
}
++j;
} else {
f[i] = f[i - 1];
}
}
return f[m];
}
}
C++
class Solution {
public:
int mincostTickets(vector<int>& days, vector<int>& costs) {
int m = days.back();
int f[m + 1];
f[0] = 0;
int valid[3] = {1, 7, 30};
for (int i = 1, j = 0; i <= m; ++i) {
if (i == days[j]) {
f[i] = INT_MAX;
for (int k = 0; k < 3; ++k) {
int c = costs[k], v = valid[k];
f[i] = min(f[i], f[max(0, i - v)] + c);
}
++j;
} else {
f[i] = f[i - 1];
}
}
return f[m];
}
};
Go
func mincostTickets(days []int, costs []int) int {
m := days[len(days)-1]
f := make([]int, m+1)
valid := [3]int{1, 7, 30}
for i, j := 1, 0; i <= m; i++ {
if i == days[j] {
f[i] = 1 << 30
for k, v := range valid {
c := costs[k]
f[i] = min(f[i], f[max(0, i-v)]+c)
}
j++
} else {
f[i] = f[i-1]
}
}
return f[m]
}
TypeScript
function mincostTickets(days: number[], costs: number[]): number {
const m = days.at(-1)!;
const f: number[] = Array(m).fill(0);
const valid: number[] = [1, 7, 30];
for (let i = 1, j = 0; i <= m; ++i) {
if (i === days[j]) {
f[i] = Infinity;
for (let k = 0; k < 3; ++k) {
const [c, v] = [costs[k], valid[k]];
f[i] = Math.min(f[i], f[Math.max(0, i - v)] + c);
}
++j;
} else {
f[i] = f[i - 1];
}
}
return f[m];
}