Minimum Number of Arrows to Burst Balloons

📋

Problem

Imagine you have a set of balloons floating along a horizontal line, each balloon occupying a segment. You want to shoot arrows vertically to burst all balloons using as few arrows as possible.

You are given an array points where points[i] = [start_i, end_i] represents the horizontal diameter of a balloon. An arrow can be shot vertically along the x-axis at any point x, bursting all balloons whose diameters intersect x. Return the minimum number of arrows that must be shot to burst all balloons.

1 ≤ points.length ≤ 10^5-2^31 ≤ start_i < end_i ≤ 2^31 - 1

Edge cases: Only one balloon → 1 arrowAll balloons non-overlapping → number of balloons arrowsAll balloons fully overlapping → 1 arrow

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IDE

def findMinArrowShots(points: list[list[int]]) -> int:public int findMinArrowShots(int[][] points)int findMinArrowShots(vector<vector<int>>& points)function findMinArrowShots(points)

def findMinArrowShots(points):
    # Write your solution here
    pass

class Solution {
    public int findMinArrowShots(int[][] points) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int findMinArrowShots(vector<vector<int>>& points) {
    // Write your solution here
    return 0;
}

function findMinArrowShots(points) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: 1Sorting intervals by start coordinate instead of end coordinate causes undercounting arrows.✅ Sort intervals by their end coordinate before applying greedy selection.

Wrong: 3Treating touching intervals as overlapping reduces arrow count incorrectly.✅ Use strict inequality (start > last arrow position) to detect overlaps.

Wrong: 0Not handling empty input correctly, returning default zero or crashing.✅ Add explicit check for empty input and return 0 arrows.

Wrong: TLEUsing nested loops for overlap checking causes O(n^2) time complexity.✅ Implement sorting by end coordinate and greedy linear scan for O(n log n) time.

✓

Test Cases

Focus on handling empty and minimal inputs correctly to avoid base case bugs.

Review your greedy approach and sorting criteria to handle tricky overlaps and boundaries.

Optimize your solution to O(n log n) by sorting and linear greedy scanning to pass large inputs.

t1_01basic

Input{"points":[[1,6],[2,8],[7,12],[10,16]]}

Expected2

⏱ Performance - must finish in 2000ms

Shoot one arrow at x = 6 to burst balloons [2,8] and [1,6], and another arrow at x = 11 to burst balloons [10,16] and [7,12].

💡 Sort balloons by their end coordinate to find overlapping groups.

💡 Use a greedy approach: pick the earliest end point to shoot an arrow and skip all balloons overlapping it.

💡 Iterate through sorted balloons, increment arrow count only when a balloon starts after the last arrow's position.

Why it failed: Incorrect arrow count due to not sorting by end or incorrect overlap check. Fix by sorting intervals by end coordinate and greedily selecting arrows at earliest possible end.

✓ Correctly counts minimum arrows using greedy scheduling by end coordinate.

t1_02basic

Input{"points":[[1,2],[3,4],[5,6],[7,8]]}

Expected4

⏱ Performance - must finish in 2000ms

All balloons are non-overlapping, so each requires a separate arrow.

💡 Check if balloons overlap by comparing start of current balloon with last arrow position.

💡 If no overlap, increment arrow count for the new balloon.

💡 Sorting by end coordinate helps identify non-overlapping intervals easily.

Why it failed: Failing to count separate arrows for non-overlapping balloons. Fix by incrementing arrow count whenever current balloon starts after last arrow position.

✓ Correctly identifies each non-overlapping balloon requires its own arrow.

t2_01edge

Input{"points":[]}

Expected0

⏱ Performance - must finish in 2000ms

No balloons means no arrows needed.

💡 Check for empty input and return 0 immediately.

💡 Handle base case where points array length is zero.

💡 Avoid processing when no balloons exist.

Why it failed: Code crashes or returns non-zero for empty input. Fix by adding a check for empty points and returning 0.

✓ Correctly returns 0 arrows for empty balloon list.

t2_02edge

Input{"points":[[5,10]]}

Expected1

⏱ Performance - must finish in 2000ms

Only one balloon requires exactly one arrow.

💡 Single balloon always needs one arrow.

💡 Check that code handles single-element input correctly.

💡 Return 1 if points length is 1.

Why it failed: Returns 0 or more than 1 for single balloon. Fix by returning 1 when only one balloon is present.

✓ Correctly returns 1 arrow for single balloon.

t2_03edge

Input{"points":[[1,5],[5,10],[10,15]]}

Expected2

⏱ Performance - must finish in 2000ms

Balloons touch at boundaries but do not overlap; each needs separate arrow.

💡 Check if touching intervals count as overlapping or not.

💡 Intervals where end equals start of next are not overlapping here.

💡 Increment arrow count for each balloon that starts after last arrow position.

Why it failed: Treating touching intervals as overlapping causing fewer arrows. Fix by using strict less than for overlap check (start > last arrow position).

✓ Correctly counts arrows when balloons only touch but do not overlap.

t3_01corner

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

Expected1

⏱ Performance - must finish in 2000ms

All smaller balloons are fully inside the first balloon, so one arrow at end=10 bursts all.

💡 Sorting by start instead of end can cause incorrect grouping.

💡 Greedy must pick balloons by earliest end to minimize arrows.

💡 Sort intervals by end coordinate, not start.

Why it failed: Greedy by start coordinate fails to minimize arrows. Fix by sorting intervals by end coordinate before greedy selection.

✓ Correctly uses end coordinate sorting to burst all balloons with one arrow.

t3_02corner

Input{"points":[[1,2],[2,3],[3,4],[4,5]]}

Expected2

⏱ Performance - must finish in 2000ms

Balloons only touch at boundaries, no overlaps; each needs separate arrow.

💡 Check if your overlap condition includes equality or strictly less.

💡 Intervals touching at boundary do not overlap for arrow bursting.

💡 Use strict inequality (start > last arrow position) to decide new arrow.

Why it failed: Counting touching intervals as overlapping reduces arrow count incorrectly. Fix by using strict inequality in overlap check.

✓ Correctly counts arrows for touching but non-overlapping balloons.

t3_03corner

Input{"points":[[-10,-5],[-7,-3],[-6,-1],[-2,0]]}

Expected2

⏱ Performance - must finish in 2000ms

Handles negative coordinates correctly; arrows at -5 and 0 burst all balloons.

💡 Negative coordinates should not affect sorting or overlap logic.

💡 Sort intervals by end coordinate regardless of sign.

💡 Apply same greedy approach for negative intervals.

Why it failed: Fails on negative coordinates due to incorrect sorting or comparison. Fix by sorting by end coordinate and comparing intervals normally.

✓ Correctly handles negative coordinates with greedy by end coordinate.

t4_01performance

Input{"_description":"n=100000 at constraint boundary - executor generates this"}

Expectednull

⏱ Performance - must finish in 2000ms

n=100000, O(n log n) sorting and O(n) greedy must complete within 2 seconds.

💡 Use efficient sorting algorithms (O(n log n)).

💡 Avoid nested loops to prevent O(n^2) time.

💡 Greedy linear scan after sorting is optimal.

Why it failed: Brute force O(n^2) approach causes TLE. Fix by sorting intervals by end coordinate and using greedy linear scan.

✓ Algorithm runs in O(n log n) time and passes performance constraints.

Practice

1. You are given a list of meeting time intervals represented as pairs of start and end times. You need to determine if a single person can attend all meetings without any overlaps. Which approach guarantees an optimal and efficient solution?

easy

A. Use dynamic programming to find the maximum number of non-overlapping intervals.

B. Sort intervals by start time and check for any overlap by comparing each interval's start with the previous interval's end.

C. Use a brute force nested loop to compare every pair of intervals for overlap.

D. Sort intervals by end time and greedily select intervals that finish earliest to maximize the number of meetings attended.

Solution

1. Step 1: Understand the problem requirement

   The problem asks if a single person can attend all meetings without overlap, so we must check if any intervals overlap.

2. Step 2: Identify the correct approach

   Sorting intervals by start time and checking if any interval starts before the previous one ends guarantees detection of overlaps efficiently.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Sorting by start time and checking overlaps is the standard approach [OK]

Hint: Sort by start time and check overlaps [OK]

Common Mistakes:

• Confusing sorting by end time with start time for overlap detection

2. What is the time complexity of the optimal car pooling solution using the difference array and prefix sum approach, given n trips and maximum location L?

medium

A. O(n * L) because we must simulate passenger count at every location for each trip

B. O(n + L) because we process each trip once and then sweep through the difference array once

C. O(n log n) due to sorting trips by start location before processing

D. O(L) because the difference array size dominates and trips are processed in constant time

Solution

1. Step 1: Analyze trip processing

   Each of the n trips updates two positions in the difference array -> O(n)

2. Step 2: Analyze prefix sum sweep

   We iterate over the difference array of size L once -> O(L)

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Sum of O(n) + O(L) operations [OK]

Hint: Difference array updates O(n), prefix sum O(L) [OK]

Common Mistakes:

• Confusing brute force O(n*L) with optimal
• Assuming sorting is required
• Ignoring prefix sum sweep cost

3. What is the time complexity of the optimal in-place merge intervals algorithm that first sorts the intervals and then merges them in one pass?

medium

A. O(n log n) due to the sorting step dominating the runtime

B. O(n) because merging is done in a single pass

C. O(n²) because each interval is compared with all others during merging

D. O(n log n) because merging requires binary searches for overlaps

Solution

1. Step 1: Identify the sorting step complexity.

   Sorting n intervals by start time takes O(n log n) time.

2. Step 2: Analyze the merging step complexity.

   Merging intervals in one pass is O(n), which is dominated by sorting.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Sorting dominates overall complexity, merging is linear. [OK]

Hint: Sorting dominates, merging is linear [OK]

Common Mistakes:

• Ignoring sorting cost
• Assuming nested comparisons cause O(n²)

4. 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

5. Suppose the problem is modified so that intervals can be reused multiple times for different queries, and queries can be negative integers. Which modification to the binary search + preprocessing approach is necessary to handle this variant correctly and efficiently?

hard

A. Preprocess intervals by length and use a segment tree to query coverage efficiently for negative and repeated queries.

B. No change needed; the original approach works for reuse and negative queries.

C. Sort queries and intervals, use a min-heap to maintain active intervals covering queries in order.

D. Use brute force checking each interval for each query since reuse breaks binary search assumptions.

Solution

1. Step 1: Identify challenges with reuse and negative queries

   Negative queries require data structures supporting arbitrary ranges; reuse means intervals must be efficiently queried multiple times.

2. Step 2: Choose appropriate data structure

   Segment tree or interval tree allows efficient coverage queries over any integer range and supports multiple queries efficiently.

3. Step 3: Why other options fail

   Min-heap approach requires sorting queries and intervals but doesn't handle negative queries well; brute force is inefficient; original approach assumes positive queries and no reuse.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Segment tree supports arbitrary queries and reuse efficiently [OK]

Hint: Segment tree handles arbitrary queries and reuse [OK]

Common Mistakes:

• Assuming original approach works unchanged
• Using min-heap without sorting queries
• Resorting to brute force unnecessarily