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.

#452

Medium

auto_awesome贪心·invariant

LeetCode 题解工作台

用最少数量的箭引爆气球

有一些球形气球贴在一堵用 XY 平面表示的墙面上。墙面上的气球记录在整数数组 points ，其中 points[i] = [x start , x end ] 表示水平直径在 x start 和 x end 之间的气球。你不知道气球的确切 y 坐标。一支弓箭可以沿着 x 轴从不同点完全垂直地…

数组贪心排序

题目描述

有一些球形气球贴在一堵用 XY 平面表示的墙面上。墙面上的气球记录在整数数组 points ，其中points[i] = [x_start, x_end] 表示水平直径在 x_start 和 x_end之间的气球。你不知道气球的确切 y 坐标。

一支弓箭可以沿着 x 轴从不同点 完全垂直 地射出。在坐标 x 处射出一支箭，若有一个气球的直径的开始和结束坐标为 x_start_，x_end_，且满足 x_start ≤ x ≤ x_end_，则该气球会被引爆 _。可以射出的弓箭的数量 没有限制 。弓箭一旦被射出之后，可以无限地前进。

给你一个数组 points ，返回引爆所有气球所必须射出的最小弓箭数 。

示例 1：

输入：points = [[10,16],[2,8],[1,6],[7,12]]
输出：2
解释：气球可以用2支箭来爆破:
-在x = 6处射出箭，击破气球[2,8]和[1,6]。
-在x = 11处发射箭，击破气球[10,16]和[7,12]。

示例 2：

输入：points = [[1,2],[3,4],[5,6],[7,8]]
输出：4
解释：每个气球需要射出一支箭，总共需要4支箭。

示例 3：

输入：points = [[1,2],[2,3],[3,4],[4,5]]
输出：2
解释：气球可以用2支箭来爆破:
- 在x = 2处发射箭，击破气球[1,2]和[2,3]。
- 在x = 4处射出箭，击破气球[3,4]和[4,5]。

提示:

1 <= points.length <= 10⁵
points[i].length == 2
-2³¹ <= x_start < x_end <= 2³¹ - 1

lightbulb

解题思路

方法一：排序 + 贪心

我们可以将所有气球按照右端点升序排序，然后从左到右遍历气球，维护当前的箭所能覆盖的最右端点 $last$ ，如果当前气球的左端点 $a$ 大于 $last$ ，说明当前箭无法覆盖当前气球，需要额外射出一支箭，那么答案加一，同时更新 $last$ 为当前气球的右端点 $b$ 。

遍历结束后，即可得到答案。

时间复杂度 $O(n \times \log n)$ ，空间复杂度 $O(\log n)$ 。其中 $n$ 为气球的数量。

相似题目：

757. 设置交集大小至少为 2

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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

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

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

warning

常见陷阱

外企场景

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

swap_horiz

进阶变体

外企场景

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

help

常见问题

外企场景

继续练习

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