983. Minimum Cost For Tickets.md

983. Minimum Cost For Tickets

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leetcode Daily Challenge on August 25th, 2020.

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Difficulty : Medium

Related Topics : Dynamic Programming

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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

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Solution

• mine
  • Java
    • Runtime: 1 ms, faster than 89.33%, Memory Usage: 39.1 MB, less than 42.02% of Java online submissions

      // O(W)time
      // O(W)space
      // W = Max(day) + 31
      public int mincostTickets(int[] days, int[] costs) {
          //dp[i] = dp[i- 1] + cost[0]   dp[i-7]+cost[1]   dp[i-30] + cost[2]
          int n = days.length;
          int max = days[n - 1] + 31;
          int[] dp = new int[max];
          int before = 0;
          for (int day : days) {
              int index = day + 30;
              getMinCost(dp, costs, before, index);
              before = index;
          }
          return dp[max - 1];
      }
      
      void getMinCost(int[] dp, int[] costs, int before, int index) {
          for (int i = before + 1; i < index; i++) {
              dp[i] = dp[before];
          }
          dp[index] = dp[index - 1] + costs[0];
          dp[index] = Math.min(dp[index], dp[index - 7] + costs[1]);
          dp[index] = Math.min(dp[index], dp[index - 30] + costs[2]);
      }
      

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• the most votes

• Dynamic Programming (Window Variant) Runtime: 0 ms, faster than 100.00%, Memory Usage: 36.9 MB, less than 97.51% of Java online submissions

  // O(N)time 
  // O(N)space   
  // N is days.length
  int[] days, costs;
  Integer[] memo;
  int[] durations = new int[]{1, 7, 30};
  public int mincostTickets(int[] days, int[] costs) {
      this.days = days;
      this.costs = costs;
      memo = new Integer[days.length];
  
      return dp(0);
  }
  
  public int dp(int i) {
      if (i >= days.length)
          return 0;
      if (memo[i] != null)
          return memo[i];
  
      int ans = Integer.MAX_VALUE;
      int j = i;
      for (int k = 0; k < 3; ++k) {
          while (j < days.length && days[j] < days[i] + durations[k])
              j++;
          ans = Math.min(ans, dp(j) + costs[k]);
      }
  
      memo[i] = ans;
      return ans;
  }
  

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