LeetCode //C - 452. Minimum Number of Arrows to Burst Balloons

452. Minimum Number of Arrows to Burst Balloons

There are some spherical balloons taped onto a flat wall that represents the XY-plane. The balloons are represented as a 2D integer array points where points[i] = [ x s t a r t , x e n d x_{start}, x_{end} xstart,xend] denotes a balloon whose horizontal diameter stretches between x s t a r t x_{start} xstart and x e n d x_{end} xend. You do not know the exact y-coordinates of the balloons.

Arrows can be shot up directly vertically (in the positive y-direction) from different points along the x-axis. A balloon with x s t a r t x_{start} xstart and x e n d x_{end} xend is burst by an arrow shot at x if x s t a r t < = x < = x e n d x_{start} <= x <= x_{end} xstart<=x<=xend. There is no limit to the number of arrows that can be shot. A shot arrow keeps traveling up infinitely, bursting any balloons in its path.

Given the array points , return the minimum number of arrows that must be shot to burst all balloons.

Example 1:

Input: points = [[10,16],[2,8],[1,6],[7,12]]
Output: 2
Explanation: 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
Explanation: 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
Explanation: 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 < = p o i n t s . l e n g t h < = 1 0 5 1 <= points.length <= 10^5 1<=points.length<=105
  • points[i].length == 2
  • − 2 31 < = x s t a r t < x e n d < = 2 31 − 1 -2^{31} <= xstart < xend <= 2^{31} - 1 −231<=xstart<xend<=231−1

From: LeetCode

Link: 452. Minimum Number of Arrows to Burst Balloons


Solution:

Ideas:

1. Problem Analysis:

The problem is essentially asking how many arrows are needed such that each arrow hits at least one balloon, and each balloon is hit by at least one arrow. An important observation here is that if an arrow is shot at some point x, it will burst all balloons whose range covers x.

2. Sorting the Balloons by End Point:

The first key idea in the solution is to sort the balloons by their ending points (i.e., x e n d x_{end} xend). The reasoning behind this is that if we shoot an arrow at the smallest available end point, we ensure that we burst as many balloons as possible that started before this end point.

The compare function helps the qsort function in sorting the balloons based on their end points.

3. Counting Arrows:

After sorting, we initialize our arrow count and set the position of the first arrow to be the end point of the first balloon.

4. Iterating Over the Balloons:

We then iterate over the rest of the balloons. For each balloon, we check its start point:

  • If the start point is less than or equal to the current arrow's position, it means this balloon can be burst by the current arrow, and we move to the next balloon.
  • If the start point is greater than the current arrow's position, it means we need a new arrow. We then increment our arrow count and set the new arrow's position to be the end point of the current balloon.

5. Return the Total Number of Arrows:

After iterating over all balloons, the arrows variable will hold the minimum number of arrows needed to burst all balloons. We return this value.

6. Handling Integer Overflow:

The initial solution had a subtraction in the compare function, which led to integer overflow for large values. We then changed the comparison logic to avoid subtraction, thereby preventing the overflow.

Code:
c 复制代码
int compare(const void* a, const void* b) {
    int end1 = (*(int**)a)[1];
    int end2 = (*(int**)b)[1];
    if (end1 < end2) return -1;
    if (end1 > end2) return 1;
    return 0;
}

int findMinArrowShots(int** points, int pointsSize, int* pointsColSize) {
    if(pointsSize == 0) {
        return 0;
    }

    // Sort the points based on the end values
    qsort(points, pointsSize, sizeof(int*), compare);

    int arrows = 1;
    int arrowPos = points[0][1];

    for(int i = 1; i < pointsSize; i++) {
        // If the start of the balloon is greater than the arrowPos, it means the current arrow can't burst this balloon
        if(points[i][0] > arrowPos) {
            arrows++;
            arrowPos = points[i][1];
        }
    }

    return arrows;
}
相关推荐
shaun20018 分钟前
华为c编程规范
c语言
MeshddY32 分钟前
(超详细)数据库项目初体验:使用C语言连接数据库完成短地址服务(本地运行版)
c语言·数据库·单片机
森焱森1 小时前
无人机三轴稳定化控制(1)____飞机的稳定控制逻辑
c语言·单片机·算法·无人机
循环过三天1 小时前
3-1 PID算法改进(积分部分)
笔记·stm32·单片机·学习·算法·pid
闪电麦坤951 小时前
数据结构:二维数组(2D Arrays)
数据结构·算法
凌肖战2 小时前
力扣网C语言编程题:快慢指针来解决 “寻找重复数”
c语言·算法·leetcode
埃菲尔铁塔_CV算法2 小时前
基于 TOF 图像高频信息恢复 RGB 图像的原理、应用与实现
人工智能·深度学习·数码相机·算法·目标检测·计算机视觉
NAGNIP3 小时前
一文搞懂FlashAttention怎么提升速度的?
人工智能·算法
Codebee3 小时前
OneCode图生代码技术深度解析:从可视化设计到注解驱动实现的全链路架构
css·人工智能·算法