#452

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Minimum Number of Arrows to Burst Balloons

Find the minimum number of arrows needed to burst all balloons by considering greedy choice and invariant validation.

Problem Statement

You are given a 2D array of balloons where each balloon is represented by two integers, xstart and xend, indicating the horizontal span of the balloon. Arrows can be shot vertically along the x-axis to burst the balloons. A balloon is burst if the arrow hits any point between its start and end positions. You are tasked with finding the minimum number of arrows required to burst all the balloons.

The balloons are represented by a list of intervals. Each arrow can burst all balloons that it passes through. You need to determine how to shoot the minimum number of arrows such that all balloons are burst. The solution involves sorting the balloons and using a greedy approach to shoot arrows where they cover the maximum possible number of balloons at once.

Examples

Example 1

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

Output: 2

The balloons can be burst by 2 arrows:

Shoot an arrow at x = 6, bursting the balloons [2,8] and [1,6].
Shoot an arrow at x = 11, bursting the balloons [10,16] and [7,12].

Example 2

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

Output: 4

One arrow needs to be shot for each balloon for a total of 4 arrows.

Example 3

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

Output: 2

The balloons can be burst by 2 arrows:

Shoot an arrow at x = 2, bursting the balloons [1,2] and [2,3].
Shoot an arrow at x = 4, bursting the balloons [3,4] and [4,5].

Constraints

1 <= points.length <= 105
points[i].length == 2
-231 <= xstart < xend <= 231 - 1

Solution Approach

Greedy Approach with Sorting

First, sort the balloons by their end positions. Then, iterate through the sorted list and shoot an arrow at the end of the first unburst balloon. This ensures that the arrow bursts as many balloons as possible, reducing the total number of arrows needed.

Validate Overlapping Intervals

After shooting an arrow, move to the next unburst balloon and check if it overlaps with the current arrow's coverage. If it does, continue without shooting another arrow. If there is no overlap, shoot another arrow at the new balloon's end position.

Efficient Arrow Placement

By focusing on the balloon's end positions rather than their start points, you ensure that each arrow is positioned optimally to burst multiple balloons in one shot. This minimizes the total number of arrows required.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n log n) due to the sorting of the balloons, followed by a linear scan of the list. The space complexity is O(1) if sorting is done in-place, or O(n) if additional space is used for sorting.

What Interviewers Usually Probe

Candidate demonstrates understanding of greedy algorithms.
Candidate can efficiently implement sorting and iteration techniques.
Candidate recognizes the importance of minimizing overlap to reduce arrow count.

Common Pitfalls or Variants

Common pitfalls

Not sorting the balloons by their end positions, which leads to suboptimal arrow placement.
Overcomplicating the problem by considering unnecessary factors like y-coordinates or specific positions within the balloon's range.
Failing to account for all balloons in one pass, resulting in unnecessary arrows.

Follow-up variants

Modify the problem to account for balloons with varying sizes.
Extend the problem to consider balloons represented in 3D space, introducing more complex geometry.
Change the arrows' movement to diagonals or allow for multi-directional shots.

FAQ

How can greedy algorithms be applied to the Minimum Number of Arrows to Burst Balloons problem?

Greedy algorithms are used in this problem by selecting arrows that burst the maximum number of balloons with each shot. Sorting the balloons by their end positions ensures that each arrow is placed optimally.

Why is sorting the balloons by end positions important in this problem?

Sorting the balloons by their end positions allows you to shoot arrows at the most efficient points, covering as many balloons as possible in one shot and minimizing the number of arrows required.

What is the time complexity of the solution?

The time complexity of the solution is O(n log n) due to the sorting step, followed by a linear scan through the balloons, which ensures optimal performance even for large inputs.

How does the problem change if there are no constraints on the number of arrows?

If there are no constraints on the number of arrows, the problem becomes trivial, as you could shoot one arrow per balloon. However, the goal is to minimize the number of arrows, making the problem more challenging.

Can this approach be used for other problems involving interval overlap?

Yes, the greedy approach used in this problem is applicable to other interval overlap problems, such as minimizing meeting room usage or scheduling tasks, where overlapping intervals need to be efficiently managed.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def findMinArrowShots(self, points: List[List[int]]) -> int:
        ans, last = 0, -inf
        for a, b in sorted(points, key=lambda x: x[1]):
            if a > last:
                ans += 1
                last = b
        return ans

Continue Practicing

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