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
SweetCode10 分钟前
裴蜀定理:整数解的奥秘
数据结构·python·线性代数·算法·机器学习
ゞ 正在缓冲99%…23 分钟前
leetcode76.最小覆盖子串
java·算法·leetcode·字符串·双指针·滑动窗口
xuanjiong24 分钟前
纯个人整理,蓝桥杯使用的算法模板day2(0-1背包问题),手打个人理解注释,超全面,且均已验证成功(附带详细手写“模拟流程图”,全网首个
算法·蓝桥杯·动态规划
小郝 小郝36 分钟前
【C语言】strstr查找字符串函数
c语言·开发语言
惊鸿.Jh43 分钟前
【滑动窗口】3254. 长度为 K 的子数组的能量值 I
数据结构·算法·leetcode
明灯L44 分钟前
《函数基础与内存机制深度剖析:从 return 语句到各类经典编程题详解》
经验分享·python·算法·链表·经典例题
碳基学AI1 小时前
哈尔滨工业大学DeepSeek公开课:探索大模型原理、技术与应用从GPT到DeepSeek|附视频与讲义免费下载方法
大数据·人工智能·python·gpt·算法·语言模型·集成学习
补三补四1 小时前
机器学习-聚类分析算法
人工智能·深度学习·算法·机器学习
独好紫罗兰1 小时前
洛谷题单3-P5718 【深基4.例2】找最小值-python-流程图重构
开发语言·python·算法
正脉科工 CAE仿真1 小时前
基于ANSYS 概率设计和APDL编程的结构可靠性设计分析
人工智能·python·算法