What is the brute force approach and complexity for Minimum Number of Arrows to Burst Balloons?

Try every subset of balloon end points as arrow spots; keep the smallest that bursts all with time complexity O(2^N · N²) and space complexity O(N).

What is the optimal approach and complexity for Minimum Number of Arrows to Burst Balloons?

Sort by end; shoot a new arrow at a balloon's end only when it starts after the last arrow with optimal time complexity O(N log N) and space complexity O(1).

How do you identify the pattern for Minimum Number of Arrows to Burst Balloons?

Look for key signals in the prompt: "an arrow bursts every balloon it touches" indicates One point can cover many intervals. Group intervals that share a point.; "minimum number of arrows" indicates Greedy choice of where each point goes: as far right as the most urgent balloon allows.; "x_start ≤ x ≤ x_end" indicates Edges count, so touching balloons share an arrow. The check is `start > arrow` for a new arrow..

Minimum Number of Arrows to Burst Balloons

Medium·LeetCode #452 Pro OnlyBasic Reinforcement

LeetCode

Problem Statement

Balloons are taped to a wall. Each balloon is given only by where it spans along the x-axis: [x_start, x_end].

You can shoot arrows straight up from any point x. An arrow at x bursts every balloon with x_start ≤ x ≤ x_end, and keeps flying.

Return the fewest arrows needed to burst all the balloons.

Note that an arrow exactly on the edge counts: an arrow at x = 2 bursts both [1, 2] and [2, 3].

Input

An array points where points[i] = [x_start, x_end].

Output

One integer: the minimum number of arrows.

Example 1

Input: points = [[10,16], [2,8], [1,6], [7,12]]

Output: 2

An arrow at x = 6 bursts [2,8] and [1,6]. An arrow at x = 11 bursts [10,16] and [7,12].

Example 2

Input: points = [[1,2], [3,4], [5,6], [7,8]]

Output: 4

No two balloons share any point, so each needs its own arrow.

Example 3

Input: points = [[1,2], [2,3], [3,4], [4,5]]

Output: 2

An arrow at x = 2 bursts [1,2] and [2,3]. An arrow at x = 4 bursts [3,4] and [4,5].

Constraints

• 1 ≤ points.length ≤ 10^5
• points[i].length == 2
• −2^31 ≤ x_start < x_end ≤ 2^31 − 1

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“an arrow bursts every balloon it touches”

→ One point can cover many intervals. Group intervals that share a point.

“minimum number of arrows”

→ Greedy choice of where each point goes: as far right as the most urgent balloon allows.

“x_start ≤ x ≤ x_end”

→ Edges count, so touching balloons share an arrow. The check is `start > arrow` for a new arrow.

Approach Complexity Ladder

Brute Force — try every set of shooting spotsO(2^N · N²)O(N)

Try every subset of balloon end points as arrow spots; keep the smallest that bursts all

Better — repeatedly shoot the balloon that ends firstO(N²)O(N)

Each round: scan for the earliest end still flying, shoot there, burst everything it touches

Optimal — sort by end, one passO(N log N)O(1)

Sort by end; shoot a new arrow at a balloon's end only when it starts after the last arrow

Brute Force Intuition

Only balloon ends are worth shooting at: slide any arrow right until it reaches the first end among the balloons it bursts, and it still bursts all of them. So collect the distinct ends, try every subset of them, and keep the smallest subset that hits every balloon.

Recognition rule for this sub-pattern:

Pick or remove intervals to optimise coverage. Sorting by end and keeping the earliest finisher leaves the most room.

Remember this · Intervals

Trigger

Input is [start, end] pairs; overlap, merge, insert, meeting rooms, free time, minimum removals.

Think

After sorting, do I only ever need to compare against the most recent interval?

Key question

Sort by start or by end — which one makes the greedy choice obvious?

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