Minimum Cost for Tickets

📋

Minimum Cost for Tickets

Edge cases:

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IDE

def mincostTickets(days: list[int], costs: list[int]) -> int:public int mincostTickets(int[] days, int[] costs)int mincostTickets(vector<int> days, vector<int> costs)function mincostTickets(days, costs)

def mincostTickets(days, costs):
    # Write your solution here
    pass

class Solution {
    public int mincostTickets(int[] days, int[] costs) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int mincostTickets(vector<int> days, vector<int> costs) {
    // Write your solution here
    return 0;
}

function mincostTickets(days, costs) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: 14Summing only 1-day passes without considering longer passes.✅ At each day, compute min cost among 1-day, 7-day, and 30-day passes covering that day.

Wrong: 0Not handling empty input or base case properly.✅ Return 0 if days array is empty.

Wrong: 9Off-by-one error in coverage calculation causing undercoverage.✅ Ensure pass covers days[i] through days[i] + duration - 1 inclusive.

Wrong: Too slow / TLEBrute force recursion without memoization or optimization.✅ Implement memoization or bottom-up DP with binary search optimization.

Wrong: Incorrect cost due to 0/1 knapsack assumptionRestricting pass purchase to once instead of unbounded times.✅ Allow multiple purchases of same pass type in DP recurrence.

✓

Test Cases

Focus on handling base cases and boundary inputs correctly.

Beware of greedy traps and 0/1 vs unbounded knapsack confusion.

Optimize your solution to handle large inputs efficiently.

t1_01basic

Input{"days":[1,4,6,7,8,20],"costs":[2,7,15]}

Expected11

⏱ Performance - must finish in 2000ms

Buy 1-day pass on day 1 (cost 2), 7-day pass on day 4 (cost 7) covering days 4-10, and 1-day pass on day 20 (cost 2). Total = 11.

💡 Consider dynamic programming to minimize cost over travel days.

💡 Use unbounded knapsack logic with passes of durations 1, 7, and 30 days.

💡 For each day, choose the minimum cost among buying 1-day, 7-day, or 30-day pass covering that day.

Why it failed: Failed basic case: likely incorrect recurrence or not considering all pass durations properly. Fix by ensuring dp[i] = min of dp[next_i] + cost for each pass duration.

✓ Correctly computes minimum cost using unbounded knapsack DP.

t1_02basic

Input{"days":[1,2,3,4,5,6,7],"costs":[2,7,15]}

Expected7

⏱ Performance - must finish in 2000ms

Buying a 7-day pass for 7 covers all days at cost 7, cheaper than multiple 1-day passes (2*7=14).

💡 Check if buying a longer duration pass is cheaper than multiple shorter passes.

💡 Use DP to compare costs of 1-day passes vs 7-day or 30-day passes.

💡 Ensure DP considers that a 7-day pass covers all days from day 1 to 7 inclusively.

Why it failed: Failed to recognize cheaper 7-day pass option; likely only summing 1-day passes. Fix by checking all pass durations at each step.

✓ Correctly identifies 7-day pass as cheapest option.

t2_01edge

Input{"days":[],"costs":[2,7,15]}

Expected0

⏱ Performance - must finish in 2000ms

No travel days means no cost.

💡 Check how your code handles empty input arrays.

💡 Base case: if no days to travel, cost should be zero.

💡 Return 0 immediately if days array is empty.

Why it failed: Fails empty input by not returning 0 or causing error. Fix by adding base case for empty days array.

✓ Correctly returns 0 cost for no travel days.

t2_02edge

Input{"days":[1],"costs":[2,7,15]}

Expected2

⏱ Performance - must finish in 2000ms

Single travel day; cheapest is 1-day pass costing 2.

💡 Test with only one travel day.

💡 Minimum cost should be the cheapest 1-day pass covering that day.

💡 Ensure DP handles single element arrays correctly.

Why it failed: Fails single element by overestimating cost or using longer passes unnecessarily. Fix by correctly handling single day base case.

✓ Correctly computes cost for single travel day.

t2_03edge

Input{"days":[1,2,3,4,5,6,7],"costs":[2,7,15]}

Expected7

⏱ Performance - must finish in 2000ms

Exact boundary test where 7-day pass covers all days exactly at cost 7.

💡 Test exact boundary where pass duration matches travel days length.

💡 Check if DP correctly uses 7-day pass to cover all days.

💡 Ensure no off-by-one errors in coverage calculation.

Why it failed: Fails exact boundary by off-by-one error in coverage range. Fix by ensuring pass covers days[i] through days[i]+duration-1 inclusively.

✓ Correctly handles exact boundary coverage.

t2_04edge

Input{"days":[1,15,30],"costs":[2,7,15]}

Expected6

⏱ Performance - must finish in 2000ms

Three 1-day passes cheaper than longer passes; tests unbounded vs 0/1 confusion.

💡 Ensure solution treats passes as unbounded (can buy multiple).

💡 Avoid 0/1 knapsack mistake where passes are considered only once.

💡 DP must allow multiple purchases of same pass type.

Why it failed: Fails unbounded knapsack by limiting pass purchase to once. Fix by allowing multiple purchases in DP recurrence.

✓ Correctly models unbounded knapsack allowing multiple passes.

t3_01corner

Input{"days":[1,2,3,4,5,6,7,8,9,10],"costs":[10,2,1]}

Expected1

⏱ Performance - must finish in 2000ms

1-day passes are expensive, 7-day pass very cheap; best to buy one 7-day pass and three 1-day passes or one 30-day pass is too expensive.

💡 Beware greedy approach that always picks cheapest pass without considering coverage.

💡 Use DP to evaluate all combinations rather than greedy selection.

💡 Check all pass options at each day to find minimal total cost.

Why it failed: Fails greedy trap by always choosing cheapest pass cost without coverage check. Fix by implementing DP that considers coverage and cost tradeoff.

✓ Avoids greedy trap and finds minimal cost correctly.

t3_02corner

Input{"days":[1,15,30],"costs":[2,7,15]}

Expected6

⏱ Performance - must finish in 2000ms

Three 1-day passes cheaper than longer passes; tests unbounded vs 0/1 confusion.

💡 Ensure solution treats passes as unbounded (can buy multiple).

💡 Avoid 0/1 knapsack mistake where passes are considered only once.

💡 DP must allow multiple purchases of same pass type.

Why it failed: Fails unbounded knapsack by limiting pass purchase to once. Fix by allowing multiple purchases in DP recurrence.

✓ Correctly models unbounded knapsack allowing multiple passes.

t3_03corner

Input{"days":[1,3,5,7,9],"costs":[3,8,20]}

Expected11

⏱ Performance - must finish in 2000ms

Tests off-by-one errors in coverage calculation; buying 1-day passes on days 1,3,5,7,9 costs 15, cheaper than 7-day or 30-day passes.

💡 Check coverage intervals carefully to avoid off-by-one errors.

💡 Verify that passes cover days from purchase day up to day + duration - 1.

💡 DP indices and coverage calculation must be precise.

Why it failed: Fails off-by-one coverage causing incorrect pass coverage and cost. Fix by adjusting coverage range to days[i] + duration - 1 inclusive.

✓ Correctly handles coverage intervals without off-by-one errors.

t4_01performance

Input{"days":[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100],"costs":[2,7,15]}

Expectednull

⏱ Performance - must finish in 2000ms

n=100, O(n^2) brute force will TLE; efficient DP with memoization or binary search optimization required.

💡 Brute force recursion is too slow for n=100.

💡 Use memoization or bottom-up DP to optimize.

💡 Consider binary search to jump indices faster.

Why it failed: TLE due to O(n^2) brute force recursion without memoization or optimization. Fix by implementing DP with memo or binary search.

✓ Efficient DP solution runs within time limits.

Practice

1. You are given a set of items, each with a weight and a value, and a knapsack with a maximum weight capacity. You want to maximize the total value of items placed in the knapsack without exceeding its capacity. Which algorithmic approach guarantees finding the optimal solution for this problem?

easy

A. A greedy algorithm that picks items with the highest value-to-weight ratio first.

B. Dynamic programming that considers including or excluding each item for every capacity up to the maximum.

C. A simple recursive approach without memoization that tries all subsets of items.

D. A breadth-first search exploring all possible combinations of items.

Solution

1. Step 1: Understand problem constraints and overlapping subproblems

   The problem requires maximizing value without exceeding capacity, which is a classic optimization problem with overlapping subproblems suitable for dynamic programming.

2. Step 2: Identify algorithmic approach that guarantees optimality

   Greedy algorithms fail because picking locally optimal items (highest value/weight) does not guarantee global optimum. Exhaustive search is correct but inefficient. Dynamic programming systematically explores all item inclusion/exclusion states with memoization, ensuring optimality.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   DP solves 0/1 Knapsack optimally by exploring all states [OK]

Hint: DP with inclusion/exclusion ensures optimal solution [OK]

Common Mistakes:

• Greedy approach works for fractional knapsack, not 0/1 knapsack.

2. You are given a list of jobs where each job has a start time, an end time, and a profit. You want to schedule jobs to maximize total profit such that no two jobs overlap. Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. Dynamic programming with jobs sorted by end time and binary search to find compatible jobs

B. Pure brute force recursion checking all subsets of jobs

C. Greedy algorithm that always picks the job with the earliest end time

D. Dynamic programming based on sorting jobs by start time and using prefix sums

Solution

1. Step 1: Understand job scheduling constraints

   The problem requires selecting non-overlapping jobs to maximize profit, which is a classic weighted interval scheduling problem.

2. Step 2: Identify the optimal approach

   Sorting jobs by end time allows binary searching for the last compatible job, enabling a DP solution that builds optimal profit incrementally.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Greedy fails on overlapping jobs with higher profit; DP with binary search handles all cases optimally [OK]

Hint: Sort by end time and use DP with binary search [OK]

Common Mistakes:

• Assuming greedy by earliest end time always works
• Sorting by start time only
• Trying brute force without pruning

3. You are given a set of positive integers and need to partition it into two subsets such that the absolute difference of their sums is minimized. Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. Greedy algorithm that sorts and assigns elements alternately to subsets

B. Dynamic programming using a subset-sum style approach to find achievable sums

C. Divide and conquer by recursively splitting the array into halves

D. Sorting and then pairing elements from opposite ends to balance sums

Solution

1. Step 1: Understand problem constraints

   The problem requires minimizing the difference between sums of two subsets, which is a classic partition problem variant.

2. Step 2: Identify algorithmic pattern

   Greedy or divide-and-conquer approaches do not guarantee minimal difference. Dynamic programming, specifically subset-sum style DP, can find all achievable sums up to total_sum, enabling minimal difference calculation.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   DP subset-sum approach guarantees optimal partition [OK]

Hint: Minimal difference requires subset-sum DP, not greedy [OK]

Common Mistakes:

• Assuming greedy or sorting suffices for minimal difference

4. The following code attempts to count the number of combinations to make change. Which line contains a subtle bug that causes it to count permutations instead of combinations?

medium

A. Line 5: inner loop iterating backwards over amounts

B. Line 4: outer loop over coins

C. Line 2: dp initialization

D. Line 6: updating dp[w] with dp[w - coin]

Solution

1. Step 1: Understand iteration order effect

   Iterating amounts backwards causes dp to count permutations, not combinations.

2. Step 2: Identify bug line

   Line 5 iterates w from amount down to coin, which breaks combination counting logic.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Forward iteration over amounts is required to count combinations correctly [OK]

Hint: Check direction of inner loop iteration [OK]

Common Mistakes:

• Using backward iteration in 1D dp
• Misplacing dp[0] initialization

5. What is the time complexity of the space-optimized bottom-up DP solution for Last Stone Weight II, where n is the number of stones and sum is the total weight of all stones?

medium

A. O(n + sum)

B. O(n^2)

C. O(n * sum * log n)

D. O(n * sum)

Solution

1. Step 1: Identify loops in the code

   Outer loop runs n times (for each stone), inner loop runs up to half the total sum (sum/2).

2. Step 2: Calculate total operations

   Each iteration updates dp array, so total operations ≈ n * (sum/2) -> O(n * sum).

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   DP complexity depends on n and sum, not n squared or log factors [OK]

Hint: DP loops over stones and sums -> O(n * sum) [OK]

Common Mistakes:

• Confusing n with sum leading to O(n^2)
• Assuming logarithmic factor due to sorting
• Ignoring that dp array size depends on sum