Minimum Number of Arrows to Burst Balloons

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Steps

setup

Check if input is empty

The algorithm first checks if the input list of balloons is empty. If it is, it returns 0 immediately.

💡 This step handles the edge case of no balloons, ensuring the algorithm doesn't proceed unnecessarily.

Line:

if not points:
    return 0

💡 The algorithm only continues if there are balloons to burst.

setup

Sort balloons by end coordinate

The balloons are sorted in ascending order by their end coordinate to prepare for the greedy approach.

💡 Sorting by end coordinate ensures we always consider balloons that finish earliest first, which is key to minimizing arrows.

Line:

points.sort(key=lambda x: x[1])

💡 Sorted balloons allow a greedy strategy to place arrows optimally.

setup

Initialize arrows count and arrow position

Set the initial number of arrows to 1 and position the first arrow at the end of the first balloon.

💡 We start with one arrow because at least one balloon exists, and placing it at the first balloon's end is optimal initially.

Line:

arrows = 1
arrow_pos = points[0][1]

💡 Starting arrow position is crucial for checking overlaps with subsequent balloons.

compare

Compare balloon 2 start with arrow position

Check if the start of the second balloon (2) is greater than the current arrow position (6).

💡 If the balloon starts after the arrow, it means the current arrow cannot burst it, so a new arrow is needed.

Line:

for start, end in points[1:]:
    if start > arrow_pos:

💡 This balloon overlaps with the current arrow's position and will be burst by it.

compare

Compare balloon 3 start with arrow position

Check if the start of the third balloon (7) is greater than the current arrow position (6).

💡 A start greater than arrow position means the current arrow cannot burst this balloon.

Line:

for start, end in points[1:]:
    if start > arrow_pos:

💡 We must increment arrows and update arrow position to burst this balloon.

insert

Shoot new arrow at balloon 3 end

Increment arrow count to 2 and update arrow position to the end of the third balloon (12).

💡 Placing the arrow at the end of the current balloon ensures it bursts as many subsequent overlapping balloons as possible.

Line:

arrows += 1
arrow_pos = end

💡 Updating arrow position is key to the greedy approach.

compare

Compare balloon 4 start with arrow position

Check if the start of the fourth balloon (10) is greater than the current arrow position (12).

💡 If the balloon starts after the arrow, a new arrow is needed; otherwise, it is burst by the current arrow.

Line:

for start, end in points[1:]:
    if start > arrow_pos:

💡 This balloon overlaps with the second arrow's position and will be burst by it.

compare

Process last balloon with current arrow

Confirm that the last balloon [10,16] is burst by the current arrow at position 12 since its start (10) <= arrow position (12).

💡 This step explicitly shows the last balloon is covered by the existing arrow, no new arrow needed.

Line:

if start > arrow_pos:
    arrows += 1
    arrow_pos = end

💡 The greedy approach efficiently covers overlapping balloons with minimal arrows.

traverse

End of iteration

All balloons have been checked against arrow positions. The total arrows used is 2.

💡 The iteration completes after all balloons are processed, finalizing the arrow count.

Line:

return arrows

💡 The greedy approach yields the minimum number of arrows needed.

reconstruct

Return final answer

The algorithm returns the total number of arrows needed, which is 2 for this input.

💡 Returning the final count completes the algorithm.

Line:

return arrows

💡 The minimum number of arrows to burst all balloons is 2.

def findMinArrowShots(points):
    # STEP 1: Check empty input
    if not points:
        return 0
    # STEP 2: Sort balloons by end coordinate
    points.sort(key=lambda x: x[1])
    # STEP 3: Initialize arrows and arrow position
    arrows = 1
    arrow_pos = points[0][1]
    # STEP 4-7: Iterate through balloons
    for start, end in points[1:]:
        # STEP 4-5: Compare start with arrow_pos
        if start > arrow_pos:
            # STEP 6: Shoot new arrow
            arrows += 1
            arrow_pos = end
    # STEP 8-9: Return total arrows
    return arrows

points = [[10,16],[2,8],[1,6],[7,12]]
print(findMinArrowShots(points))  # Output: 2

📊

Minimum Number of Arrows to Burst Balloons - Watch the Algorithm Execute, Step by Step

Watching the algorithm step-by-step reveals how sorting by end coordinate enables a greedy approach to minimize arrows, which is hard to grasp from code alone.

check_empty

0

1

2

3

sort

0

1

2

3

initialize

arrow_pos

0

i

1

2

3

Result: 1

compare

arrow_pos

0

i

1

2

3

Result: 1

compare

arrow_pos

0

1

i

2

3

Result: 1

insert

0

1

i

2

3

Result: 2

compare

0

1

arrow_pos

2

i

3

Result: 2

compare

0

1

arrow_pos

2

i

3

Result: 2

return

0

1

2

3

Result: 2

final_answer

0

1

2

3

Result: 2

Key Takeaways

✓ Sorting balloons by their end coordinate enables a greedy approach that guarantees the minimum number of arrows.

This insight is difficult to see from code alone because the importance of sorting is conceptual and visual.

✓ Placing an arrow at the end of the earliest finishing balloon covers all overlapping balloons starting before that point.

Visualizing this shows why the arrow position updates only when a balloon starts after the current arrow.

✓ Each time a balloon starts after the current arrow position, a new arrow is needed, incrementing the count.

Seeing the pointer comparisons and arrow position updates clarifies when and why decisions branch.

Practice

1. Given multiple employees' schedules with intervals representing their busy times, which algorithmic approach guarantees finding all common free time intervals where no employee is busy?

easy

A. Greedy interval scheduling that picks earliest finishing intervals

B. Dynamic programming to find maximum non-overlapping intervals

C. Sweep line / event processing that tracks interval start and end events

D. Brute force checking every time point across all intervals

Solution

1. Step 1: Understand problem requires identifying gaps where no intervals overlap

   Greedy scheduling or DP focus on selecting intervals, not gaps between them.

2. Step 2: Sweep line processes all start/end events in order, tracking active intervals to find free gaps

   This approach efficiently detects when no employee is busy, yielding correct free times.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Sweep line tracks active intervals and finds free gaps [OK]

Hint: Sweep line tracks interval endpoints to find free gaps [OK]

Common Mistakes:

• Assuming greedy or DP can find free gaps directly

2. The following code attempts to implement the optimal car pooling solution. Which line contains a subtle bug that can cause incorrect capacity checks?

def carPooling(trips, capacity):
    max_location = 0
    for _, start, end in trips:
        max_location = max(max_location, end)
    diff = [0] * (max_location + 1)
    for num, start, end in trips:
        diff[start] += num
        diff[end] += num  # Bug here
    current_passengers = 0
    for i in range(max_location + 1):
        current_passengers += diff[i]
        if current_passengers > capacity:
            return false
    return true

medium

A. Line that increments diff[end] by num instead of decrementing

B. Line that updates max_location in the first loop

C. Line that increments diff[start] by num

D. Line that checks if current_passengers exceeds capacity

Solution

1. Step 1: Understand difference array updates

   Passengers boarding at start location increase count; passengers leaving at end location must decrease count.

2. Step 2: Identify incorrect update

   Line incrementing diff[end] by num adds passengers instead of removing them, causing incorrect capacity calculation.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Increment at end location breaks capacity tracking [OK]

Hint: End location must decrement passengers, not increment [OK]

Common Mistakes:

• Incrementing instead of decrementing at end
• Forgetting to update difference array
• Checking capacity before summing passengers

3. Examine the following buggy code snippet for the minimum interval problem. Which line contains the subtle bug that can cause incorrect results on some inputs?

def min_interval_binary_search(intervals, queries):
    intervals.sort(key=lambda x: x[1] - x[0] + 1)
    lengths = [end - start for start, end in intervals]
    starts = [start for start, end in intervals]
    ends = [end for start, end in intervals]

    def covers(query, length):
        idx = bisect.bisect_right(lengths, length)
        for i in range(idx):
            if starts[i] <= query <= ends[i]:
                return True
        return False

    res = []
    max_len = lengths[-1] if lengths else 0
    for q in queries:
        left, right = 1, max_len
        ans = -1
        while left <= right:
            mid = (left + right) // 2
            if covers(q, mid):
                ans = mid
                right = mid - 1
            else:
                left = mid + 1
        res.append(ans)
    return res

medium

A. Line 15: max_len = lengths[-1] if lengths else 0

B. Line 2: intervals.sort(key=lambda x: x[1] - x[0] + 1)

C. Line 9: for i in range(idx):

D. Line 3: lengths = [end - start for start, end in intervals]

Solution

1. Step 1: Check interval length calculation

   Line 3 calculates lengths as end - start, missing the +1 needed for inclusive intervals.

2. Step 2: Impact of bug

   Incorrect lengths cause wrong sorting order and coverage checks, leading to wrong minimal interval lengths.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Correct interval length must be end - start + 1 [OK]

Hint: Check interval length calculation carefully [OK]

Common Mistakes:

• Off-by-one in length calculation
• Not sorting intervals
• Incorrect binary search boundaries

4. Suppose the car pooling problem is modified so that trips can overlap and passengers can be picked up and dropped off multiple times (i.e., trips can reuse the same passengers). Which of the following algorithmic changes correctly adapts the solution to handle this variant efficiently?

hard

A. Use a segment tree or binary indexed tree to efficiently update and query passenger counts over intervals

B. Switch to a priority queue to process events in order and track current passengers dynamically

C. Use the same difference array approach but multiply passenger counts by the number of trips to simulate reuse

D. Modify the difference array to allow negative passenger counts and track net changes carefully

Solution

1. Step 1: Understand reuse complexity

   Reusing passengers means overlapping intervals can increase and decrease counts multiple times, requiring efficient range updates and queries.

2. Step 2: Identify suitable data structure

   Segment trees or binary indexed trees support efficient interval updates and queries, handling overlapping intervals with reuse.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Segment trees handle dynamic interval sums efficiently [OK]

Hint: Reuse -> need data structure for interval updates [OK]

Common Mistakes:

• Multiplying counts breaks logic
• Priority queue alone can't handle reuse efficiently
• Allowing negative counts without structure causes errors

5. Suppose intervals can be reused multiple times (i.e., you can merge overlapping intervals repeatedly, possibly reusing intervals). Which modification to the merge intervals algorithm correctly handles this scenario?

hard

A. Use the same sorting + one pass merge approach; repeated intervals will be merged naturally.

B. Sort intervals and repeatedly apply the merge pass until no changes occur, which may take multiple passes.

C. Use a brute force nested loop approach to merge intervals until no overlaps remain.

D. Use a graph-based approach to find connected components of overlapping intervals and merge each component.

Solution

1. Step 1: Understand that reusing intervals means overlaps can form complex connected groups.

   Simple one-pass merge assumes intervals appear once and sorted order suffices.

2. Step 2: Recognize that overlapping intervals form connected components in a graph.

   Building a graph where intervals are nodes and edges represent overlaps allows merging all connected intervals correctly.

3. Step 3: Evaluate options.

   Use the same sorting + one pass merge approach; repeated intervals will be merged naturally. fails because one pass merge assumes no reuse. Sort intervals and repeatedly apply the merge pass until no changes occur, which may take multiple passes. is inefficient and may not terminate quickly. Use a brute force nested loop approach to merge intervals until no overlaps remain. is brute force and inefficient. Use a graph-based approach to find connected components of overlapping intervals and merge each component. correctly models the problem and merges all connected intervals.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   Graph connected components capture all overlapping intervals including reuse. [OK]

Hint: Model reuse as graph connected components [OK]

Common Mistakes:

• Assuming one pass merge works with reuse
• Using repeated merges without termination guarantee