Dynamic programming Minimum Cost for ticked Leetcode
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 <= days.length <= 365
1 <= days[i] <= 365
days
is in strictly increasing order.costs.length == 3
1 <= costs[i] <= 1000
Accepted
44,224
Submissions
74,271
class Solution { public int mincostTickets(int[] days, int[] costs) { boolean travel[] = new boolean[366]; for(int i = 0; i<days.length;i++ ){ travel[days[i]] =true; // System.out.println("i"+" --"+days[i]+" + "+travel[days[i]]); } System.out.println(days[days.length-1]); int dp[] = new int[366]; dp[0]=0; for(int i = 1; i<=365; i++){ if(!travel[i]){ dp[i] =dp[i-1]; continue; } dp[i] =Math.min(Math.min( dp[Math.max(0, i-1)]+costs[0],dp[Math.max(0, i-7)]+costs[1]),dp[Math.max(0, i-30)]+costs[2]); } return dp[days[days.length-1]]; } }