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
YuforiaCode14 分钟前
第十二届蓝桥杯 2021 C/C++组 直线
c语言·c++·蓝桥杯
知来者逆33 分钟前
计算机视觉——速度与精度的完美结合的实时目标检测算法RF-DETR详解
图像处理·人工智能·深度学习·算法·目标检测·计算机视觉·rf-detr
阿让啊38 分钟前
C语言中操作字节的某一位
c语言·开发语言·数据结构·单片机·算法
এ᭄画画的北北38 分钟前
力扣-160.相交链表
算法·leetcode·链表
拾忆-eleven1 小时前
C语言实战:用Pygame打造高难度水果消消乐游戏
c语言·python·pygame
爱研究的小陈2 小时前
Day 3:数学基础回顾——线性代数与概率论在AI中的核心作用
算法
渭雨轻尘_学习计算机ing2 小时前
二叉树的最大宽度计算
算法·面试
再睡一夏就好2 小时前
Linux常见工具如yum、vim、gcc、gdb的基本使用,以及编译过程和动静态链接的区别
linux·服务器·c语言·c++·笔记
BB_CC_DD2 小时前
四. 以Annoy算法建树的方式聚类清洗图像数据集,一次建树,无限次聚类搜索,提升聚类搜索效率。(附完整代码)
深度学习·算法·聚类
embedded_w3 小时前
U8G2在PC端模拟(C语言版本)
c语言