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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 1day pass is sold for
costs[0]
dollars,  a 7day pass is sold for
costs[1]
dollars, and  a 30day pass is sold for
costs[2]
dollars.
The passes allow that many days of consecutive travel.
 For example, if we get a 7day pass on day
2
, then we can travel for7
days: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 1day pass for costs[0] = 2, which covered day 1. On day 3, you bought a 7day pass for costs[1] = 7, which covered days 3, 4, ..., 9. On day 20, you bought a 1day 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 30day pass for costs[2] = 15 which covered days 1, 2, ..., 30. On day 31, you bought a 1day 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 <= 365
1 <= days[i] <= 365
days
is in strictly increasing order.costs.length == 3
1 <= costs[i] <= 1000
Solutions

class Solution { private static final int[] T = new int[] {1, 7, 30}; private int[] costs; private int[] days; private int[] f; private int n; public int mincostTickets(int[] days, int[] costs) { n = days.length; f = new int[n]; this.costs = costs; this.days = days; Arrays.fill(f, 1); return dfs(0); } private int dfs(int i) { if (i >= n) { return 0; } if (f[i] != 1) { return f[i]; } int res = Integer.MAX_VALUE; for (int k = 0; k < 3; ++k) { int j = lowerBound(days, days[i] + T[k]); res = Math.min(res, costs[k] + dfs(j)); } f[i] = res; return res; } private int lowerBound(int[] days, int x) { int left = 0, right = days.length; while (left < right) { int mid = (left + right) >> 1; if (days[mid] >= x) { right = mid; } else { left = mid + 1; } } return left; } }

class Solution { public: vector<int> t = {1, 7, 30}; vector<int> days; vector<int> costs; vector<int> f; int n; int mincostTickets(vector<int>& days, vector<int>& costs) { n = days.size(); this>days = days; this>costs = costs; f.assign(n, 1); return dfs(0); } int dfs(int i) { if (i >= n) return 0; if (f[i] != 1) return f[i]; int res = INT_MAX; for (int k = 0; k < 3; ++k) { int j = lower_bound(days.begin(), days.end(), days[i] + t[k])  days.begin(); res = min(res, costs[k] + dfs(j)); } f[i] = res; return res; } };

class Solution: def mincostTickets(self, days: List[int], costs: List[int]) > int: @cache def dfs(i): if i >= len(days): return 0 res = inf for c, d in zip(costs, [1, 7, 30]): j = bisect_left(days, days[i] + d) res = min(res, c + dfs(j)) return res return dfs(0)

func mincostTickets(days []int, costs []int) int { t := []int{1, 7, 30} n := len(days) f := make([]int, n) for i := range f { f[i] = 1 } var dfs func(i int) int dfs = func(i int) int { if i >= n { return 0 } if f[i] != 1 { return f[i] } res := 0x3f3f3f3f for k, c := range costs { j := lowerBound(days, days[i]+t[k]) res = min(res, c+dfs(j)) } f[i] = res return res } return dfs(0) } func lowerBound(arr []int, x int) int { left, right := 0, len(arr) for left < right { mid := (left + right) >> 1 if arr[mid] >= x { right = mid } else { left = mid + 1 } } return left }

function mincostTickets(days: number[], costs: number[]): number { const n = days.length, m = days[n  1] + 1; const [a, b, c] = costs; let dp = new Array(m).fill(0); for (let i = 1; i < m; i++) { let x = days.includes(i) ? dp[i  1] + a : dp[i  1]; let y = (i > 7 ? dp[i  7] : dp[0]) + b; let z = (i > 30 ? dp[i  30] : dp[0]) + c; dp[i] = Math.min(x, y, z); } return dp[m  1]; }