How do you solve Minimum Cost For Tickets on LeetCode?

Minimum Cost For Tickets (LeetCode #983) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Dynamic programming helps avoid redundant calculations.

What is the brute force solution for Minimum Cost For Tickets?

The brute force approach for Minimum Cost For Tickets is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible combination of ticket purchases for each travel day. It calculates the total cost for each combination and keeps track of the minimum cost found. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Cost For Tickets?

Minimum Cost For Tickets uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Minimum Cost For Tickets?

Explain your thought process clearly while coding. Consider edge cases, like no travel days or all days being travel days. Practice writing clean and efficient code.

What are common mistakes when solving Minimum Cost For Tickets?

Not considering all ticket types when calculating costs. Failing to handle days that are not travel days.

#983

Minimum Cost For Tickets

Medium

ArrayDynamic ProgrammingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses dynamic programming to build up the minimum cost for each day based on previously computed results. This avoids redundant calculations and efficiently finds the minimum cost.

⚙️

Algorithm

3 steps

1Step 1: Create a DP array where dp[i] represents the minimum cost to travel up to day i.
2Step 2: Iterate through each travel day and update the DP array based on the cost of each ticket type.
3Step 3: Return the value in the DP array corresponding to the last travel day.

solution.py13 lines

1def mincostTickets(days, costs):
2    dp = [0] * 366
3    travel_days = set(days)
4    for day in range(1, 366):
5        if day in travel_days:
6            dp[day] = min(
7                dp[day - 1] + costs[0],
8                dp[max(0, day - 7)] + costs[1],
9                dp[max(0, day - 30)] + costs[2]
10            )
11        else:
12            dp[day] = dp[day - 1]
13    return dp[365]

ℹ

Complexity note: The time complexity is O(n) because we iterate through each day of the year once, updating the DP array. The space complexity is O(n) due to the DP array storing costs for each day.

1Dynamic programming helps avoid redundant calculations.
2Understanding the problem constraints (days within 1-365) is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.