Problem Description

You're planning train travel for a year and have a list of specific days when you'll need to travel. These travel days are given as an integer array days, where each day is a number from 1 to 365.

There are three types of train passes available for purchase:

• A 1-day pass costs costs[0] dollars and covers travel for 1 consecutive day
• A 7-day pass costs costs[1] dollars and covers travel for 7 consecutive days
• A 30-day pass costs costs[2] dollars and covers travel for 30 consecutive days

When you buy a pass on a specific day, it covers that day and the specified number of consecutive days following it. For example, if you buy a 7-day pass on day 2, you can travel on days 2, 3, 4, 5, 6, 7, and 8.

Your goal is to find the minimum total cost needed to cover all the travel days in your list. You need to strategically choose which passes to buy and when to buy them so that every day in your days array is covered by at least one pass, while minimizing the total amount spent.

The solution uses dynamic programming with memoization. The function dfs(i) calculates the minimum cost to cover all travel days starting from index i in the sorted days array. For each position, it considers buying each of the three pass types and uses binary search to find the next uncovered travel day after the current pass expires. The algorithm recursively calculates costs for all possibilities and returns the minimum.

Intuition

The key insight is that we need to make a decision at each travel day: which type of pass should we buy to minimize the total cost? This naturally leads to a dynamic programming approach where we explore all possible choices and select the optimal one.

Think of it this way: when we're standing at a particular travel day, we have three options - buy a 1-day, 7-day, or 30-day pass. Each choice will cover some future travel days, and then we'll need to solve the same problem again for the remaining uncovered days. This recursive structure suggests we can break down the problem into smaller subproblems.

For each travel day at index i, we ask: "What's the minimum cost to cover all travel days from index i to the end?" To answer this, we:

1. Try buying a 1-day pass (costs costs[0]) which covers only the current day
2. Try buying a 7-day pass (costs costs[1]) which might cover multiple upcoming travel days
3. Try buying a 30-day pass (costs costs[2]) which might cover even more travel days

After buying any pass, we need to find the next uncovered travel day. Since our travel days are sorted, we can use binary search to efficiently find the first day that isn't covered by our current pass. For example, if we're on day 5 and buy a 7-day pass, we need to find the first travel day that's >= day 12 (5 + 7).

The recursive formula becomes: dfs(i) = min(cost_of_pass + dfs(next_uncovered_day)) for each pass type. We use memoization to avoid recalculating the same subproblems multiple times, as different combinations of passes might lead us to the same remaining travel days.

The base case is when we've covered all travel days (i >= n), at which point the cost is 0 since no more passes are needed.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution implements a top-down dynamic programming approach with memoization and binary search optimization.

Implementation Details:

1. Main Function Structure: We define a recursive function dfs(i) that calculates the minimum cost to cover all travel days starting from index i in the days array. The answer to our problem is dfs(0).

2. Memoization with @cache: The @cache decorator is used to automatically memoize the results of dfs(i). This prevents redundant calculations when the same subproblem is encountered multiple times through different paths.

3. Base Case:

   if i >= n:
       return 0

   When i reaches or exceeds the total number of travel days n, we've covered all required days, so the additional cost is 0.

4. Recursive Case: For each position i, we try all three pass types:

   ans = inf
   for c, v in zip(costs, valid):
       j = bisect_left(days, days[i] + v)
       ans = min(ans, c + dfs(j))

   • We initialize ans to infinity to track the minimum cost
   • valid = [1, 7, 30] represents the duration of each pass type
   • We iterate through each pass type using zip(costs, valid)

5. Binary Search Optimization:

   j = bisect_left(days, days[i] + v)

   For each pass type with duration v, we use bisect_left to find the index j of the first travel day that would NOT be covered by this pass. Specifically:

   • If we're at days[i] and buy a pass valid for v days
   • The pass covers days from days[i] to days[i] + v - 1
   • We search for the first day >= days[i] + v in the sorted days array
   • This gives us the starting index for the next subproblem

6. Cost Calculation: For each pass type, we calculate:

   • Current pass cost: c
   • Cost for remaining days: dfs(j)
   • Total cost: c + dfs(j)

   We keep track of the minimum among all three options.

Time Complexity: O(n × log n) where n is the length of the days array. Each of the n states is computed once due to memoization, and each state performs 3 binary searches, each taking O(log n) time.

Space Complexity: O(n) for the memoization cache and recursion stack.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Example Input:

• days = [1, 4, 6, 7, 8, 20]
• costs = [2, 7, 15] (1-day: $2, 7-day:$7, 30-day: $15)

Step-by-step execution of dfs(i):

Initial call: dfs(0) (starting at day 1)

• Current day: days[0] = 1
• Option 1: Buy 1-day pass ($2)
  • Covers day 1 only
  • Next uncovered day: bisect_left([1,4,6,7,8,20], 1+1=2) → index 1 (day 4)
  • Cost: $2 + dfs(1)
• Option 2: Buy 7-day pass ($7)
  • Covers days 1-7
  • Next uncovered day: bisect_left([1,4,6,7,8,20], 1+7=8) → index 4 (day 8)
  • Cost: $7 + dfs(4)
• Option 3: Buy 30-day pass ($15)
  • Covers days 1-30
  • Next uncovered day: bisect_left([1,4,6,7,8,20], 1+30=31) → index 6 (beyond array)
  • Cost: $15 + `dfs(6)` =$15 + 0 = $15

Exploring Option 1: dfs(1) (starting at day 4)

• Current day: days[1] = 4
• Option 1: Buy 1-day pass ($2)
  • Next uncovered: index 2 (day 6)
  • Cost: $2 + dfs(2)
• Option 2: Buy 7-day pass ($7)
  • Covers days 4-10
  • Next uncovered: index 5 (day 20)
  • Cost: $7 + dfs(5)
• Option 3: Buy 30-day pass ($15)
  • Covers all remaining days
  • Cost: $15 + 0 =$15

Continuing dfs(2) (starting at day 6)

• Current day: days[2] = 6
• Option 1: Buy 1-day pass ($2)
  • Next uncovered: index 3 (day 7)
  • Cost: $2 + dfs(3)
• Option 2: Buy 7-day pass ($7)
  • Covers days 6-12
  • Next uncovered: index 5 (day 20)
  • Cost: $7 + dfs(5)

And dfs(3) (starting at day 7)

• Current day: days[3] = 7
• Option 1: Buy 1-day pass ($2)
  • Next uncovered: index 4 (day 8)
  • Cost: $2 + dfs(4)

Then dfs(4) (starting at day 8)

• Current day: days[4] = 8
• Option 1: Buy 1-day pass ($2)
  • Next uncovered: index 5 (day 20)
  • Cost: $2 + dfs(5)

Finally dfs(5) (starting at day 20)

• Current day: days[5] = 20
• Option 1: Buy 1-day pass ($2)
  • Next uncovered: index 6 (end of array)
  • Cost: $2 + 0 =$2
• Option 2: Buy 7-day pass ($7)
  • Cost: $7 + 0 =$7
• Option 3: Buy 30-day pass ($15)
  • Cost: $15 + 0 =$15
• Minimum: $2

Working backwards with memoized results:

• dfs(5) = 2
• dfs(4) = 2 + 2 = 4
• dfs(3) = 2 + 4 = 6
• dfs(2) = min(2 + 6, 7 + 2) = min(8, 9) = 8
• dfs(1) = min(2 + 8, 7 + 2, 15) = min(10, 9, 15) = 9
• dfs(0) = min(2 + 9, 7 + 4, 15) = min(11, 11, 15) = 11

Final answer: $11

The optimal strategy is to buy a 7-day pass on day 1 (covering days 1,4,6,7) and a 1-day pass on day 8, and another 1-day pass on day 20, for a total of $7 +$2 + $2 =$11.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × log n)

The time complexity is determined by the recursive DFS function with memoization. There are n possible states (one for each index i from 0 to n-1), and each state is computed at most once due to the @cache decorator. For each state, we iterate through 3 ticket options (1-day, 7-day, and 30-day passes), and for each option, we perform a binary search using bisect_left to find the next index, which takes O(log n) time. Therefore, the overall time complexity is O(n × 3 × log n) = O(n × log n).

Space Complexity: O(n)

The space complexity consists of two main components:

1. The recursion call stack, which can go up to depth n in the worst case when we process each day sequentially
2. The memoization cache from the @cache decorator, which stores at most n computed results (one for each possible starting index)

Both contribute O(n) space, resulting in an overall space complexity of O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Off-by-One Error in Pass Coverage Calculation

The Pitfall: A common mistake is incorrectly calculating when a pass expires. Many developers mistakenly use:

j = bisect_left(days, days[i] + v - 1)  # WRONG!

This searches for the last day covered by the pass, not the first uncovered day.

Why It's Wrong:

• A 7-day pass bought on day 2 covers days 2-8 (inclusive)
• The first uncovered day is day 9, which is 2 + 7 = 9
• Using days[i] + v - 1 would give us day 8, which is still covered

The Correct Approach:

j = bisect_left(days, days[i] + v)  # CORRECT

This correctly finds the index of the first day that needs a new pass.

2. Using bisect_right Instead of bisect_left

The Pitfall:

j = bisect_right(days, days[i] + v - 1)  # Subtle bug!

Why It's Wrong: If days[i] + v - 1 exists in the array, bisect_right would skip over it and potentially miss calculating costs for some days. Consider:

• days = [1, 4, 6], currently at days[0] = 1 with a 3-day pass
• Pass covers days 1, 2, 3
• bisect_right(days, 3) returns index 1 (correctly)
• But if days = [1, 3, 4, 6] and we use bisect_right(days, 3), it returns index 2, skipping day 4

The Fix: Always use bisect_left with days[i] + v to consistently find the first uncovered day.

3. Forgetting to Handle Edge Cases in Binary Search

The Pitfall: Not considering that binary search might return an index equal to len(days):

j = bisect_left(days, days[i] + v)
next_cost = costs[pass_type] + dfs(j)  # j might be >= len(days)

Why It Matters: When the pass covers all remaining travel days, bisect_left returns len(days). The base case if i >= n: return 0 handles this correctly, but developers sometimes add unnecessary bounds checking that can introduce bugs.

The Solution: Trust the base case to handle i >= n. The algorithm naturally handles this scenario without additional checks.

4. Incorrect Pass Duration Values

The Pitfall:

valid = [0, 6, 29]  # WRONG! Zero-indexed thinking

Why It's Wrong: The pass durations represent the number of days covered, not array indices:

• 1-day pass covers 1 day (not 0)
• 7-day pass covers 7 days (not 6)
• 30-day pass covers 30 days (not 29)

The Correct Values:

valid = [1, 7, 30]  # Actual number of days covered

5. Not Memoizing the Recursive Function

The Pitfall: Forgetting the @cache decorator leads to exponential time complexity:

def dfs(i):  # Missing @cache decorator
    if i >= n:
        return 0
    # ... rest of the code

Why It's Critical: Without memoization, the same subproblems are solved multiple times. For example, different combinations of passes might lead to the same remaining days to cover, causing redundant calculations.

The Solution: Always use @cache or implement manual memoization with a dictionary to store computed results.