Maximum Profit in Job Scheduling

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Steps

setup

Sort jobs by end time

Jobs are sorted by their end times to process them in chronological order of completion.

💡 Sorting by end time ensures we consider jobs in a way that respects scheduling constraints.

Line:

jobs = sorted(zip(startTime, endTime, profit), key=lambda x: x[1])

💡 Sorting is crucial to efficiently find compatible jobs later.

setup

Initialize dp and ends arrays

Create dp array to store max profit up to each job and ends array for binary search.

💡 dp array holds max profit results; ends array helps find compatible jobs quickly.

Line:

n = len(jobs)
ends = [job[1] for job in jobs]
dp = [0] * n

💡 Prepares data structures for efficient DP computation.

fill_cells

Process job 0: (1,3,20)

For the first job, no previous compatible job exists; dp[0] is profit of job 0.

💡 First job sets base case for dp array.

Line:

start_i, end_i, profit_i = jobs[0]
idx = bisect.bisect_right(ends, start_i - 1) - 1
# idx = -1
take = profit_i + (dp[idx] if idx >= 0 else 0)
skip = 0
dp[0] = max(take, skip)

💡 When no compatible job exists, max profit is just current job's profit.

fill_cells

Process job 1: (2,5,20)

Find last compatible job for job 1; idx = -1 means none; compare taking or skipping job 1.

💡 Binary search finds no compatible job before job 1 starts.

Line:

start_i, end_i, profit_i = jobs[1]
idx = bisect.bisect_right(ends, start_i - 1) - 1
# idx = -1
take = profit_i + (dp[idx] if idx >= 0 else 0)
skip = dp[0]
dp[1] = max(take, skip)

💡 Skipping or taking job 1 yields same profit; dp[1] remains 20.

fill_cells

Process job 2: (4,6,70)

Find last compatible job for job 2; idx = 0 (job 0 ends at 3, before job 2 starts at 4).

💡 Binary search finds job 0 compatible with job 2.

Line:

start_i, end_i, profit_i = jobs[2]
idx = bisect.bisect_right(ends, start_i - 1) - 1
# idx = 0
take = profit_i + dp[idx]
skip = dp[1]
dp[2] = max(take, skip)

💡 Taking job 2 plus compatible job 0 yields higher profit.

fill_cells

Process job 3: (6,9,60) - Find compatible job

Find last compatible job for job 3; idx = 2 (job 2 ends at 6, job 3 starts at 6).

💡 Jobs ending exactly when next starts are compatible.

Line:

start_i, end_i, profit_i = jobs[3]
idx = bisect.bisect_right(ends, start_i - 1) - 1
# idx = 2

💡 Identifying compatible job is key before deciding to take or skip.

fill_cells

Process job 3: (6,9,60) - Decide take or skip

Calculate take = profit of job 3 + dp[idx], skip = dp[2]; choose max.

💡 Compare profit taking job 3 plus compatible jobs vs skipping it.

Line:

take = profit_i + dp[idx]
skip = dp[2]
dp[3] = max(take, skip)

💡 Taking job 3 plus compatible jobs yields max profit so far.

fill_cells

Process job 4: (3,10,100) - Find compatible job

Find last compatible job for job 4; idx = 0 (job 0 ends at 3, job 4 starts at 3).

💡 Job 4 overlaps with jobs 1, 2, 3; only job 0 is compatible.

Line:

start_i, end_i, profit_i = jobs[4]
idx = bisect.bisect_right(ends, start_i - 1) - 1
# idx = 0

💡 Identifying compatible job is essential before decision.

fill_cells

Process job 4: (3,10,100) - Decide take or skip

Calculate take = profit of job 4 + dp[idx], skip = dp[3]; choose max.

💡 Compare profit taking job 4 plus compatible jobs vs skipping it.

Line:

take = profit_i + dp[idx]
skip = dp[3]
dp[4] = max(take, skip)

💡 Skipping job 4 yields higher profit than taking it.

reconstruct

Final result read from dp array

The maximum profit is the last value in dp array, representing best schedule profit.

💡 The dp array accumulates max profit; last cell is final answer.

Line:

return dp[-1]

💡 dp[-1] = 150 is the maximum profit achievable.

from typing import List
import bisect

def jobScheduling(startTime: List[int], endTime: List[int], profit: List[int]) -> int:
    jobs = sorted(zip(startTime, endTime, profit), key=lambda x: x[1])  # STEP 1
    n = len(jobs)  # STEP 2
    ends = [job[1] for job in jobs]  # STEP 2
    dp = [0] * n  # STEP 2

    for i in range(n):  # STEP 3-9
        start_i, end_i, profit_i = jobs[i]  # STEP i
        idx = bisect.bisect_right(ends, start_i - 1) - 1  # STEP i
        take = profit_i + (dp[idx] if idx >= 0 else 0)  # STEP i
        skip = dp[i - 1] if i > 0 else 0  # STEP i
        dp[i] = max(take, skip)  # STEP i

    return dp[-1]  # STEP 10

if __name__ == '__main__':
    print(jobScheduling([1,2,3,4,6], [3,5,10,6,9], [20,20,100,70,60]))  # Expected 150

📊

Maximum Profit in Job Scheduling - Watch the Algorithm Execute, Step by Step

Watching each step reveals how the algorithm balances taking or skipping jobs based on compatibility and profit, making the DP logic clear without needing to read code.

Initializing dp array

i\w	0	1	2	3	4
i=0	?	?	?	?	?

Job 1 (1,3,20)

Initializing dp array

i\w	0	1	2	3	4
i=0	?	?	?	?	?

dp unfilled

Item 0 - wt:3 val:20

i\w	0	1	2	3	4
i=0	20	?	?	?	?

dp[0] = 20 (take job 0)

Item 1 - wt:5 val:20

i\w	0	1	2	3	4
i=0	20	20	?	?	?

dp[1] = 20 (skip or take job 1)

Item 2 - wt:6 val:70

i\w	0	1	2	3	4
i=0	20	20	90	?	?

dp[2] = 90 (take job 2 + dp[0])

Item 3 - wt:9 val:60

i\w	0	1	2	3	4
i=0	20	20	90	?	?

Compatible job idx = 2

Item 3 - wt:9 val:60

i\w	0	1	2	3	4
i=0	20	20	90	150	?

dp[3] = 150 (take job 3 + dp[2])

Item 4 - wt:10 val:100

i\w	0	1	2	3	4
i=0	20	20	90	150	?

Compatible job idx = 0

Item 4 - wt:10 val:100

i\w	0	1	2	3	4
i=0	20	20	90	150	150

dp[4] = 150 (skip job 4)

Item 4 - wt:10 val:100

i\w	0	1	2	3	4
i=0	20	20	90	150	150

Answer: 150

Key Takeaways

✓ Sorting jobs by end time is essential for efficient compatibility checks.

Without sorting, finding compatible jobs would be inefficient and complex.

✓ The dp array stores the maximum profit achievable up to each job, reflecting cumulative decisions.

This shows how dynamic programming builds solutions incrementally.

✓ Binary search quickly finds the last compatible job, enabling O(n log n) complexity.

This optimization is key to making the algorithm efficient and practical.

Practice

1. Consider the following code snippet implementing the minimum cost for tickets problem. What is the value of dp[0] after the loop completes for the input days = [1,4,6] and costs = [2,7,15]?

easy

A. 6

B. 7

C. 9

D. 4

Solution

1. Step 1: Trace dp from the end

   dp[3] = 0 (base case). For i=2 (day=6): - 1-day ticket: cost=2 + dp[3]=2 - 7-day ticket: cost=7 + dp[3]=7 - 30-day ticket: cost=15 + dp[3]=15 Minimum = 2 -> dp[2]=2

2. Step 2: Calculate dp[1] and dp[0]

   i=1 (day=4): - 1-day: cost=2 + dp[2]=4 - 7-day: cost=7 + dp[3]=7 - 30-day: cost=15 + dp[3]=15 Minimum=4 -> dp[1]=4 i=0 (day=1): - 1-day: cost=2 + dp[1]=6 - 7-day: cost=7 + dp[3]=7 - 30-day: cost=15 + dp[3]=15 Minimum=6 -> dp[0]=6

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   dp[0] correctly computed as 6 [OK]

Hint: Trace dp from end to start carefully [OK]

Common Mistakes:

• Off-by-one in dp indexing
• Miscomputing next_i with bisect
• Confusing costs and dp sums

2. Given the following code, what is the output when coins = [1, 2, 5] and amount = 5?

def count_ways_space_opt(coins, amount):
    dp = [0] * (amount + 1)
    dp[0] = 1
    for coin in coins:
        for w in range(coin, amount + 1):
            dp[w] += dp[w - coin]
    return dp[amount]

coins = [1, 2, 5]
amount = 5
print(count_ways_space_opt(coins, amount))

easy

A. 4

B. 7

C. 5

D. 6

Solution

1. Step 1: Trace dp array updates for each coin

   Start dp = [1,0,0,0,0,0]. After coin=1, dp = [1,1,1,1,1,1]. After coin=2, dp = [1,1,2,2,3,3]. After coin=5, dp = [1,1,2,2,3,6].

2. Step 2: Confirm dp[5] value

   dp[5] = 6, representing 6 ways to make amount 5 with coins [1,2,5].

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Manual dp tracing matches output 6 [OK]

Hint: DP accumulates counts incrementally per coin [OK]

Common Mistakes:

• Off-by-one in dp indexing
• Miscounting after last coin iteration
• Confusing permutations with combinations

3. Consider the following buggy code for lastStoneWeightII. Which line contains the subtle bug that causes incorrect results on some inputs?

medium

A. Line 4: dp[0] = true is missing

B. Line 7: Outer loop over stones

C. Line 8: Inner loop iterates backwards

D. Line 10: Return statement calculation

Solution

1. Step 1: Identify dp initialization

   dp[0] must be true to represent sum zero achievable with no stones; missing this means no sums are marked achievable.

2. Step 2: Consequence of missing dp[0] = true

   Without dp[0] = true, dp array remains false, so no sums can be formed, causing the function to fail to find any valid partition.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Initializing dp[0] is critical base case for subset sum DP [OK]

Hint: dp[0] = true is base case for subset sums [OK]

Common Mistakes:

• Forgetting dp[0] initialization
• Iterating forward in inner loop causing double counting
• Miscomputing return value

4. Suppose the problem is modified so that each string can be chosen multiple times (unbounded usage). Which change to the dynamic programming approach correctly adapts the solution?

hard

A. Iterate dp array forwards to allow multiple counts of the same string

B. Iterate dp array backwards as before to avoid counting duplicates

C. Use recursion without memoization to handle repeated choices

D. Increase dp dimensions to track usage count of each string

Solution

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

   Unbounded knapsack allows multiple uses of the same item, so dp updates must allow reuse within the same iteration.

2. Step 2: Adjust dp iteration order

   Iterating dp forwards enables using updated states multiple times, correctly counting repeated strings.

3. Step 3: Confirm correctness

   Backward iteration prevents reuse in the same iteration, which is incorrect for unbounded usage.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Forward iteration enables multiple item usage [OK]

Hint: Unbounded knapsack uses forward dp iteration [OK]

Common Mistakes:

• Continuing backward iteration from 0/1 knapsack
• Adding extra dp dimensions unnecessarily

5. Suppose the problem is modified so that each element in the array can be used multiple times (unlimited reuse) to form k equal sum subsets. Which of the following modifications to the DP with bitmask tabulation approach correctly adapts to this change?

hard

A. Use backtracking with memoization on subsets but do not use bitmasking since reuse breaks subset uniqueness

B. Keep the same bitmask DP but allow setting dp[next_mask] multiple times without restriction

C. Replace bitmask DP with a classic unbounded knapsack DP that tracks sums without bitmasking

D. Modify the bitmask DP to reset dp states after each full subset sum is reached to allow reuse

Solution

1. Step 1: Understand reuse impact

   Allowing unlimited reuse breaks the uniqueness assumption of subsets represented by bitmasks.

2. Step 2: Identify suitable DP approach

   Classic unbounded knapsack DP tracks sums without bitmasking, correctly handling reuse.

3. Step 3: Evaluate other options

   Bitmask DP relies on unique element usage; modifying it without removing bitmasking leads to incorrect states.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Unbounded knapsack DP handles reuse correctly [OK]

Hint: Bitmask DP assumes unique usage; reuse needs unbounded knapsack DP [OK]

Common Mistakes:

• Trying to reuse bitmask DP without removing uniqueness assumption
• Ignoring that reuse breaks subset partitioning logic