Minimum Cost for Tickets

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Steps

setup

Initialize dp array

Create a dp array of length n+1 initialized with zeros, where n is the number of travel days. dp[n] = 0 means no cost after the last day.

💡 Initializing dp with zeros sets the base case: no cost when no days remain to cover.

Line:

dp = [0] * (n + 1)

💡 dp array length and base case are established for bottom-up computation.

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Start filling dp from last day index 5

Set i = 5, the last index in days. We will compute dp[5], the cost to cover days from day 20 onward.

💡 Filling dp from the last day backward ensures all future costs are known when computing current dp.

Line:

for i in range(n - 1, -1, -1):

💡 We begin with the last day to use dp[next_i] values that are already computed.

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Evaluate 1-day ticket at day 20

Check cost of buying a 1-day ticket on day 20. Find next uncovered day index after day 20 + 1 = 21 using binary search.

💡 Binary search finds the next day not covered by the ticket, so we know which dp value to add.

Line:

next_i = bisect.bisect_left(days, days[i] + d, i, n)

💡 No more days after day 20, so cost is ticket cost + 0.

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Evaluate 7-day ticket at day 20

Check cost of buying a 7-day ticket on day 20. Next uncovered day is after day 27, binary search returns index 6 (end).

💡 Longer tickets may cover more days but cost more; we compare all options.

Line:

next_i = bisect.bisect_left(days, days[i] + d, i, n)

💡 7-day ticket covers day 20 and beyond, cost is 7.

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Evaluate 30-day ticket at day 20

Check cost of buying a 30-day ticket on day 20. Next uncovered day after day 50 is index 6 (end).

💡 Even longer tickets may not be cost-effective for sparse days.

Line:

next_i = bisect.bisect_left(days, days[i] + d, i, n)

💡 30-day ticket is most expensive here.

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Choose minimum cost for dp[5]

Minimum cost among ticket options is 2 (1-day ticket). Set dp[5] = 2.

💡 Choosing the minimum cost ensures optimal solution for subproblem starting at day 20.

Line:

dp[i] = ans

💡 dp[5] now holds the minimal cost to cover days from index 5 onward.

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Compute dp[4] for day 8

Set i = 4 for day 8. We will compute dp[4], the cost to cover days from day 8 onward.

💡 Moving backward to previous day index to fill dp array.

Line:

for i in range(n - 1, -1, -1):

💡 We continue filling dp from right to left.

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Evaluate 1-day ticket at day 8

Check cost of 1-day ticket at day 8. Next uncovered day after 9 is index 5 (day 20).

💡 Binary search finds next uncovered day index to add dp cost.

Line:

next_i = bisect.bisect_left(days, days[i] + d, i, n)

💡 1-day ticket covers only day 8, then pay dp[5] for remaining days.

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Evaluate 7-day ticket at day 8

Check cost of 7-day ticket at day 8. Next uncovered day after 15 is index 6 (end).

💡 Longer ticket covers days 8-14, skipping days 20.

Line:

next_i = bisect.bisect_left(days, days[i] + d, i, n)

💡 7-day ticket covers all remaining days except day 20.

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Evaluate 30-day ticket at day 8

Check cost of 30-day ticket at day 8. Next uncovered day after 38 is index 6 (end).

💡 30-day ticket covers all remaining days.

Line:

next_i = bisect.bisect_left(days, days[i] + d, i, n)

💡 30-day ticket is most expensive but covers all days.

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Choose minimum cost for dp[4]

Minimum cost is 4 (1-day ticket + dp[5]). Set dp[4] = 4.

💡 Choosing minimum cost ensures optimal subproblem solution.

Line:

dp[i] = ans

💡 dp[4] now holds minimal cost from day 8 onward.

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Compute dp[3] for day 7

Set i = 3 for day 7. Compute dp[3], cost to cover days from day 7 onward.

💡 Continue backward filling dp array.

Line:

for i in range(n - 1, -1, -1):

💡 We proceed to next day index.

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Evaluate 1-day ticket at day 7

Check cost of 1-day ticket at day 7. Next uncovered day after 8 is index 4 (day 8).

💡 Binary search finds next uncovered day index.

Line:

next_i = bisect.bisect_left(days, days[i] + d, i, n)

💡 1-day ticket covers day 7 only, then pay dp[4].

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Evaluate 7-day ticket at day 7

Check cost of 7-day ticket at day 7. Next uncovered day after 14 is index 5 (day 20).

💡 7-day ticket covers days 7-13, skipping day 20.

Line:

next_i = bisect.bisect_left(days, days[i] + d, i, n)

💡 7-day ticket covers more days but costs more.

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Evaluate 30-day ticket at day 7

Check cost of 30-day ticket at day 7. Next uncovered day after 37 is index 6 (end).

💡 30-day ticket covers all remaining days.

Line:

next_i = bisect.bisect_left(days, days[i] + d, i, n)

💡 30-day ticket is most expensive option.

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Choose minimum cost for dp[3]

Minimum cost is 6 (1-day ticket + dp[4]). Set dp[3] = 6.

💡 Choosing minimum cost ensures optimal subproblem solution.

Line:

dp[i] = ans

💡 dp[3] now holds minimal cost from day 7 onward.

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Compute dp[0] for day 1 (summary step)

Jumping ahead to dp[0] for brevity: compute minimal cost starting from day 1 by evaluating all ticket options similarly.

💡 This step summarizes the final dp computation to reach the answer.

Line:

for i in range(n - 1, -1, -1): ... dp[i] = ans

💡 dp[0] holds the minimal total cost to cover all travel days.

reconstruct

Read final answer from dp[0]

The minimal cost to cover all travel days is dp[0] = 11. This is the output of the algorithm.

💡 Reading dp[0] gives the answer to the original problem.

Line:

return dp[0]

💡 The dp array stores all subproblem solutions culminating in the final answer.

import bisect

def mincostTickets(days, costs):
    n = len(days)  # STEP 1
    durations = [1,7,30]
    dp = [0] * (n + 1)  # STEP 1

    for i in range(n - 1, -1, -1):  # STEP 2,7,12,17
        ans = float('inf')
        for j, d in enumerate(durations):  # STEP 3-5,8-10,13-15
            cost = costs[j]
            next_day = days[i] + d
            next_i = bisect.bisect_left(days, next_day, i, n)  # STEP 3,8,13
            ans = min(ans, cost + dp[next_i])  # STEP 3,8,13
        dp[i] = ans  # STEP 6,11,16,17

    return dp[0]  # STEP 18

if __name__ == '__main__':
    print(mincostTickets([1,4,6,7,8,20], [2,7,15]))  # Output: 11

📊

Minimum Cost for Tickets - Watch the Algorithm Execute, Step by Step

Watching the dp array fill step-by-step reveals how the algorithm balances ticket durations and costs to minimize total expense, which is hard to grasp from code alone.

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Key Takeaways

✓ The dp array stores the minimal cost to cover all travel days starting from each index, computed backwards.

This backward filling and dependency on future dp values is hard to visualize from code alone.

✓ Binary search efficiently finds the next uncovered day index after buying a ticket, enabling quick dp lookups.

Understanding this optimization clarifies how the algorithm avoids redundant checks.

✓ Comparing all ticket options at each step reveals the tradeoff between ticket duration and cost, leading to the optimal choice.

Seeing each option evaluated step-by-step makes the decision process concrete.

Practice

1. Consider the following Python code for the Equal Partition problem. What is the value of dp[target] after processing the input [1, 5, 11, 5]?

def canPartition(nums):
    total = sum(nums)
    if total % 2 != 0:
        return False
    target = total // 2
    dp = [False] * (target + 1)
    dp[0] = True

    for num in nums:
        for w in range(target, num - 1, -1):
            dp[w] = dp[w] or dp[w - num]

    return dp[target]

print(canPartition([1,5,11,5]))

easy

A. true

B. false

C. null (error occurs)

D. Depends on order of iteration

Solution

1. Step 1: Calculate total and target

   The total sum is 1 + 5 + 11 + 5 = 22, which is even, so target = 11.

2. Step 2: Trace dp array updates for each num

   After processing all nums, dp[11] becomes True because subset {11} or {5,5,1} sums to 11.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Input is classic example where partition exists; dp[target] is True [OK]

Hint: Sum is even and dp[target] tracks subset sums [OK]

Common Mistakes:

• Confusing dp[target] with dp[target-1]
• Forgetting to iterate backwards in dp update

2. What is the time complexity of the space-optimized bottom-up dynamic programming solution for counting the number of ways to make change with n coins and amount W?

medium

A. O(n * W) because for each coin we iterate over all amounts up to W

B. O(n + W) because we iterate over coins and amounts separately

C. O(W^n) due to nested loops over coins and amounts

D. O(n * W) plus O(W) auxiliary space for the dp array

Solution

1. Step 1: Identify loops and their ranges

   Outer loop runs n times (coins), inner loop runs up to W (amount), so time is O(n * W).

2. Step 2: Consider space usage

   DP array size is W+1, so auxiliary space is O(W). Total complexity includes both time and space, but time complexity is the main focus here.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Time O(n*W) matches DP approach [OK]

Hint: Time is product of coins and amount; space is dp array size [OK]

Common Mistakes:

• Confusing time with space complexity
• Forgetting auxiliary space for dp array
• Mistaking exponential complexity for DP

3. Examine the following code snippet which attempts to solve the problem. Identify the line containing the subtle bug that causes incorrect results on some inputs.

medium

A. Line 7: zeros = s.count('0')

B. Line 10: for j in range(0, n - ones + 1):

C. Line 9: for i in range(0, m - zeros + 1): # iterating forwards

D. Line 11: dp[i + zeros][j + ones] = max(dp[i + zeros][j + ones], 1 + dp[i][j])

Solution

1. Step 1: Understand dp iteration direction

   To avoid counting the same string multiple times, dp must be updated backwards (from high to low indices).

2. Step 2: Identify bug in iteration

   Iterating forwards over i and j causes dp states to be reused within the same iteration, leading to overcounting.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Backward iteration is required for 0/1 knapsack variants [OK]

Hint: Forward iteration in 0/1 knapsack dp causes double counting [OK]

Common Mistakes:

• Forgetting to iterate dp backwards
• Miscounting zeros and ones

4. Suppose the problem changes so that you can use each item an unlimited number of times (unbounded knapsack). Which modification to the space-optimized bottom-up DP code correctly solves this variant?

hard

A. Change the inner loop to iterate forwards from weights[i] up to W.

B. Add memoization to the recursive solution to avoid recomputation.

C. Change the inner loop to iterate backwards from W down to weights[i].

D. Use a greedy approach selecting items with the highest value-to-weight ratio repeatedly.

Solution

1. Step 1: Understand difference between 0/1 and unbounded knapsack

   In unbounded knapsack, items can be chosen multiple times, so dp[w] can be updated using dp[w - weights[i]] from the same iteration.

2. Step 2: Identify correct iteration order

   Iterating forwards from weights[i] to W allows reuse of updated dp values within the same iteration, enabling multiple counts of the same item.

3. Step 3: Why other options fail

   Backward iteration prevents multiple usage in one iteration (wrong for unbounded). Memoization helps but is not the direct fix. Greedy fails for 0/1 and unbounded knapsack except fractional case.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Forward iteration enables unlimited item reuse [OK]

Hint: Unbounded knapsack requires forward iteration in dp updates [OK]

Common Mistakes:

• Using backward iteration from 0/1 knapsack causes incorrect results.

5. Suppose the coin change problem is modified so that each coin can only be used at most once (0/1 knapsack variant). Which of the following changes to the bottom-up DP approach correctly solves this variant?

hard

A. Use a 2D dp array where dp[i][j] represents minimum coins to make amount j using first i coins, iterating coins outer and amounts inner

B. Sort coins and greedily pick largest coins without repetition until amount is reached

C. Keep the same 1D dp array and iterate coins inside the amount loop as before (unbounded usage)

D. Use the same 1D dp but iterate amounts outer and coins inner, updating dp[j] only if dp[j - coin] was from previous iteration

Solution

1. Step 1: Understand the 0/1 usage constraint

   Each coin can be used once, so we must track usage per coin, requiring 2D dp.

2. Step 2: Identify correct DP structure

   2D dp with dp[i][j] = min coins to make j using first i coins ensures no reuse; iterate coins outer, amounts inner.

3. Step 3: Why other options fail

   Keep the same 1D dp array and iterate coins inside the amount loop as before (unbounded usage) allows unlimited reuse; B is greedy and incorrect; D misuses 1D dp which is for unbounded usage.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   0/1 knapsack requires 2D dp to track coin usage [OK]

Hint: 0/1 knapsack needs 2D dp to prevent reuse [OK]

Common Mistakes:

• Using unbounded knapsack DP for 0/1 problem
• Assuming greedy works for minimum coins