Car Pooling

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Steps

setup

Find maximum location

We scan all trips to find the maximum end location to size the difference array correctly.

💡 Knowing the maximum location ensures the difference array covers all relevant points where passengers change.

Line:

max_location = 0
for _, start, end in trips:
    max_location = max(max_location, end)

💡 The maximum location is 7, so the difference array must have length 8 (0 to 7).

setup

Initialize difference array

Create a difference array of zeros with length max_location + 1 to track passenger changes.

💡 The difference array will record increments and decrements of passengers at trip start and end locations.

Line:

diff = [0] * (max_location + 1)

💡 The difference array is initialized with zeros for all locations 0 through 7.

fill_cells

Add passengers for first trip start

Add 2 passengers at location 1, the start of the first trip.

💡 Incrementing at the start location marks passengers getting on the car.

Line:

diff[start] += num  # diff[1] += 2

💡 Difference array at index 1 changes from 0 to 2.

fill_cells

Subtract passengers for first trip end

Subtract 2 passengers at location 5, the end of the first trip.

💡 Decrementing at the end location marks passengers getting off the car.

Line:

diff[end] -= num  # diff[5] -= 2

💡 Difference array at index 5 changes from 0 to -2.

fill_cells

Add passengers for second trip start

Add 3 passengers at location 3, the start of the second trip.

💡 Incrementing at location 3 marks 3 passengers boarding the car.

Line:

diff[start] += num  # diff[3] += 3

💡 Difference array at index 3 changes from 0 to 3.

fill_cells

Subtract passengers for second trip end

Subtract 3 passengers at location 7, the end of the second trip.

💡 Decrementing at location 7 marks 3 passengers leaving the car.

Line:

diff[end] -= num  # diff[7] -= 3

💡 Difference array at index 7 changes from 0 to -3.

setup

Initialize current passengers count

Set current_passengers to zero before computing prefix sums.

💡 We will accumulate passenger changes to find total passengers at each location.

Line:

current_passengers = 0

💡 Current passengers start at zero before scanning locations.

traverse

Add passengers at location 0

Add diff[0] (0) to current_passengers, total remains 0.

💡 Start prefix sum from location 0 to accumulate passenger counts.

Line:

current_passengers += diff[i]  # i=0

💡 No passengers board or leave at location 0.

traverse

Add passengers at location 1

Add diff[1] (2) to current_passengers, total becomes 2.

💡 Passengers board at location 1, increasing total passengers.

Line:

current_passengers += diff[i]  # i=1

💡 Passenger count is now 2, within capacity 4.

traverse

Add passengers at location 2

Add diff[2] (0) to current_passengers, total remains 2.

💡 No passenger change at location 2, total stays the same.

Line:

current_passengers += diff[i]  # i=2

💡 Passenger count remains 2, still within capacity.

traverse

Add passengers at location 3

Add diff[3] (3) to current_passengers, total becomes 5.

💡 3 passengers board at location 3, increasing total passengers.

Line:

current_passengers += diff[i]  # i=3

💡 Passenger count exceeds capacity at location 3, so car pooling is not possible.

prune

Return false due to capacity breach

Since passengers exceed capacity at location 3, the function returns false immediately.

💡 Early termination saves unnecessary computation once capacity is exceeded.

Line:

if current_passengers > capacity:
    return False

💡 The algorithm efficiently detects failure without scanning all locations.

def carPooling(trips, capacity):
    max_location = 0  # STEP 1
    for _, start, end in trips:
        max_location = max(max_location, end)  # STEP 1
    diff = [0] * (max_location + 1)  # STEP 2
    for num, start, end in trips:
        diff[start] += num  # STEP 3 & 5
        diff[end] -= num    # STEP 4 & 6
    current_passengers = 0  # STEP 7
    for i in range(max_location + 1):
        current_passengers += diff[i]  # STEP 8-11
        if current_passengers > capacity:  # STEP 11
            return False  # STEP 12
    return True

if __name__ == '__main__':
    print(carPooling([[2,1,5],[3,3,7]], 4))  # False
    print(carPooling([[2,1,5],[3,3,7]], 5))  # True

📊

Car Pooling - Watch the Algorithm Execute, Step by Step

Watching each update and prefix sum calculation reveals how the algorithm efficiently tracks overlapping intervals without checking every trip repeatedly.

move_right

Result: 7

setup

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Result: 2

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compare

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Result: 5

prune

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Result: false

Key Takeaways

✓ Difference arrays let us efficiently track net passenger changes at trip boundaries without iterating over all trips repeatedly.

This insight is hard to see from code alone because the difference array concept is abstract until you see incremental updates.

✓ Prefix sums over the difference array reveal the total passengers at each location, showing overlaps clearly.

Visualizing prefix sums step-by-step clarifies how overlapping intervals accumulate passenger counts.

✓ Early termination when capacity is exceeded saves computation and shows the algorithm's efficiency.

Seeing the exact location and passenger count that causes failure helps understand the pruning logic.

Practice

1. You are given a list of meeting time intervals consisting of start and end times. Your task is to find the minimum number of conference rooms required so that all meetings can be held without overlap. Which of the following approaches guarantees an optimal solution with efficient time complexity?

easy

A. Sort intervals by start time and use a min-heap to track the earliest ending meeting to reuse rooms efficiently.

B. Use a brute force approach checking every pair of intervals for overlap and counting maximum overlaps.

C. Use dynamic programming to find the maximum number of non-overlapping intervals and subtract from total intervals.

D. Sort intervals by end time and greedily assign rooms without tracking ongoing meetings.

Solution

1. Step 1: Understand the problem requires tracking overlapping intervals to find minimum rooms.

   Brute force checks all pairs but is inefficient; greedy by end time alone doesn't track concurrent overlaps.

2. Step 2: Recognize that sorting by start time and using a min-heap to track earliest ending meeting allows reusing rooms optimally.

   This approach efficiently manages room allocation by freeing rooms as meetings end.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Min-heap approach is standard optimal solution for this problem [OK]

Hint: Min-heap tracks earliest end to reuse rooms efficiently [OK]

Common Mistakes:

• Assuming greedy by end time alone suffices
• Thinking brute force is efficient enough
• Confusing DP for interval scheduling with room allocation

2. You are given a list of time intervals representing booked meeting rooms. Your task is to combine all overlapping intervals into the minimum number of non-overlapping intervals. Which algorithmic approach guarantees an optimal and efficient solution for this problem?

easy

A. Use a greedy approach that always merges the earliest starting interval with the next one without sorting.

B. Sort intervals by start time and then merge overlapping intervals in a single pass.

C. Use dynamic programming to find the maximum number of non-overlapping intervals and then merge accordingly.

D. Check every pair of intervals with nested loops to merge overlaps until no overlaps remain.

Solution

1. Step 1: Understand the problem requires merging overlapping intervals efficiently.

   Sorting intervals by their start time ensures that overlapping intervals are adjacent, enabling a single pass merge.

2. Step 2: Evaluate each option's approach.

   Use a greedy approach that always merges the earliest starting interval with the next one without sorting. fails because without sorting, intervals may be merged incorrectly or missed. Use dynamic programming to find the maximum number of non-overlapping intervals and then merge accordingly. is unrelated as DP is not needed here. Check every pair of intervals with nested loops to merge overlaps until no overlaps remain. is brute force with O(n²) time, inefficient for large inputs. Sort intervals by start time and then merge overlapping intervals in a single pass. correctly sorts and merges in O(n log n) time.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Sorting + one pass merge is the classic optimal pattern for merging intervals. [OK]

Hint: Sort intervals first, then merge in one pass [OK]

Common Mistakes:

• Trying to merge without sorting
• Using nested loops leading to O(n²) time

3. What is the time complexity of the binary search + preprocessing approach for the Minimum Interval to Include Each Query problem, given n intervals and m queries?

medium

A. O(n log n + m n log n)

B. O(n log n + m log n * n)

C. O((n + m) log n)

D. O(n * m)

Solution

1. Step 1: Analyze preprocessing

   Sorting intervals by length takes O(n log n).

2. Step 2: Analyze per query cost

   For each query, binary search over lengths is O(log n), but coverage check scans up to O(n) intervals, so O(n log n) per query.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Overall complexity is O(n log n + m n log n) due to coverage scanning [OK]

Hint: Coverage check inside binary search causes O(n) scan per query [OK]

Common Mistakes:

• Assuming coverage check is O(log n) ignoring linear scan
• Confusing n and m in complexity
• Forgetting sorting cost

4. Suppose the intervals list can contain overlapping intervals initially (not guaranteed sorted or disjoint), and you want to insert a new interval and return the merged list. Which modification to the optimal approach is necessary to handle this variant correctly?

hard

A. Insert the new interval in the correct position without sorting, then merge only intervals overlapping with the new interval.

B. Append the new interval, then sort and merge all intervals including the original overlapping ones.

C. Skip sorting and merge intervals only if they overlap with the new interval to improve efficiency.

D. Use a balanced tree data structure to insert and merge intervals dynamically without sorting.

Solution

1. Step 1: Understand initial intervals may overlap

   Since intervals are not guaranteed sorted or disjoint, merging must consider all intervals, not just those overlapping newInterval.

2. Step 2: Modify approach to handle all overlaps

   Appending newInterval and sorting all intervals ensures correct order, then merging all intervals handles existing overlaps and new overlaps caused by insertion.

3. Step 3: Why other options fail

   Options B and C assume partial merging which misses existing overlaps; D is overcomplicated and not required.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Sorting and merging all intervals handles arbitrary overlaps [OK]

Hint: Sort and merge all intervals when initial list may overlap [OK]

Common Mistakes:

• Assuming initial intervals are sorted and disjoint
• Trying partial merges only around new interval
• Skipping sorting leads to incorrect merges

5. Suppose the interval lists can contain intervals with negative numbers and intervals within the same list may overlap. Which modification to the original two pointers approach correctly handles this scenario?

hard

A. Preprocess each list by merging overlapping intervals before applying the two pointers intersection algorithm.

B. Modify the two pointers algorithm to skip intervals with negative numbers, as they cannot intersect positively.

C. Use the brute force nested loop approach since two pointers rely on non-overlapping intervals within each list.

D. Sort the combined intervals from both lists and then find intersections by scanning sequentially.

Solution

1. Step 1: Identify problem with overlapping intervals

   Two pointers assume each list is sorted and non-overlapping internally; overlapping intervals break this assumption.

2. Step 2: Preprocess by merging intervals

   Merging each list's overlapping intervals produces sorted, non-overlapping lists suitable for two pointers.

3. Step 3: Apply two pointers intersection

   After preprocessing, the original algorithm works correctly even with negative numbers.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Preprocessing restores assumptions needed for efficient intersection [OK]

Hint: Merge intervals first to restore non-overlapping property [OK]

Common Mistakes:

• Ignoring internal overlaps
• Skipping negative intervals incorrectly
• Switching to brute force unnecessarily