983. Minimum Cost For Tickets LeetCode Solution

Last updated: February 21, 2025

In this guide, you will get 983. Minimum Cost For Tickets LeetCode Solution with the best time and space complexity. The solution to Minimum Cost For Tickets problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Minimum Cost For Tickets solution in C++
Minimum Cost For Tickets solution in Java
Minimum Cost For Tickets solution in Python
Additional Resources

983. Minimum Cost For Tickets LeetCode Solution image

Problem Statement of Minimum Cost For Tickets

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 for 7 days: 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.

Constraints:

1 <= days.length <= 365
1 <= days[i] <= 365
days is in strictly increasing order.
costs.length == 3
1 <= costs[i] <= 1000

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n)

983. Minimum Cost For Tickets LeetCode Solution in C++

class Solution {
 public:
  int mincostTickets(vector<int>& days, vector<int>& costs) {
    int ans = 0;
    queue<pair<int, int>> last7;
    queue<pair<int, int>> last30;

    for (const int day : days) {
      while (!last7.empty() && last7.front().first + 7 <= day)
        last7.pop();
      while (!last30.empty() && last30.front().first + 30 <= day)
        last30.pop();
      last7.emplace(day, ans + costs[1]);
      last30.emplace(day, ans + costs[2]);
      ans = min({ans + costs[0], last7.front().second, last30.front().second});
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

983. Minimum Cost For Tickets LeetCode Solution in Java

class Solution {
  public int mincostTickets(int[] days, int[] costs) {
    int ans = 0;
    Queue<Pair<Integer, Integer>> last7 = new ArrayDeque<>(); // [day, cost]
    Queue<Pair<Integer, Integer>> last30 = new ArrayDeque<>();

    for (final int day : days) {
      while (!last7.isEmpty() && last7.peek().getKey() + 7 <= day)
        last7.poll();
      while (!last30.isEmpty() && last30.peek().getKey() + 30 <= day)
        last30.poll();
      last7.offer(new Pair<>(day, ans + costs[1]));
      last30.offer(new Pair<>(day, ans + costs[2]));
      ans = Math.min(ans + costs[0], Math.min(last7.peek().getValue(), last30.peek().getValue()));
    }

    return ans;
  }
}
// code provided by PROGIEZ

983. Minimum Cost For Tickets LeetCode Solution in Python

class Solution:
  def mincostTickets(self, days: list[int], costs: list[int]) -> int:
    ans = 0
    last7 = collections.deque()
    last30 = collections.deque()

    for day in days:
      while last7 and last7[0][0] + 7 <= day:
        last7.popleft()
      while last30 and last30[0][0] + 30 <= day:
        last30.popleft()
      last7.append([day, ans + costs[1]])
      last30.append([day, ans + costs[2]])
      ans = min(ans + costs[0], last7[0][1], last30[0][1])

    return ans
 # code by PROGIEZ

Additional Resources

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