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Co-Authored-By: Antim-IWP <203163676+Antim-IWP@users.noreply.github.com>
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Author: 3 people

Solution/452. Minimum Number of Arrows to Burst Balloons/readme.md

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+---
+comments: true
+difficulty: Medium
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/0400-0499/0452.Minimum%20Number%20of%20Arrows%20to%20Burst%20Balloons/README_EN.md
+tags:
+    - Greedy
+    - Array
+    - Sorting
+---
+
+<!-- problem:start -->
+
+# [452. Minimum Number of Arrows to Burst Balloons](https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons)
+
+[中文文档](/solution/0400-0499/0452.Minimum%20Number%20of%20Arrows%20to%20Burst%20Balloons/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>There are some spherical balloons taped onto a flat wall that represents the XY-plane. The balloons are represented as a 2D integer array <code>points</code> where <code>points[i] = [x<sub>start</sub>, x<sub>end</sub>]</code> denotes a balloon whose <strong>horizontal diameter</strong> stretches between <code>x<sub>start</sub></code> and <code>x<sub>end</sub></code>. You do not know the exact y-coordinates of the balloons.</p>
+
+<p>Arrows can be shot up <strong>directly vertically</strong> (in the positive y-direction) from different points along the x-axis. A balloon with <code>x<sub>start</sub></code> and <code>x<sub>end</sub></code> is <strong>burst</strong> by an arrow shot at <code>x</code> if <code>x<sub>start</sub> &lt;= x &lt;= x<sub>end</sub></code>. There is <strong>no limit</strong> to the number of arrows that can be shot. A shot arrow keeps traveling up infinitely, bursting any balloons in its path.</p>
+
+<p>Given the array <code>points</code>, return <em>the <strong>minimum</strong> number of arrows that must be shot to burst all balloons</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> points = [[10,16],[2,8],[1,6],[7,12]]
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> 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].
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> points = [[1,2],[3,4],[5,6],[7,8]]
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> One arrow needs to be shot for each balloon for a total of 4 arrows.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> points = [[1,2],[2,3],[3,4],[4,5]]
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> 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].
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= points.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>points[i].length == 2</code></li>
+	<li><code>-2<sup>31</sup> &lt;= x<sub>start</sub> &lt; x<sub>end</sub> &lt;= 2<sup>31</sup> - 1</code></li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+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
+```
+
+#### Java
+
+```java
+class Solution {
+    public int findMinArrowShots(int[][] points) {
+        // 直接 a[1] - b[1] 可能会溢出
+        Arrays.sort(points, Comparator.comparingInt(a -> a[1]));
+        int ans = 0;
+        long last = -(1L << 60);
+        for (var p : points) {
+            int a = p[0], b = p[1];
+            if (a > last) {
+                ++ans;
+                last = b;
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int findMinArrowShots(vector<vector<int>>& points) {
+        sort(points.begin(), points.end(), [](vector<int>& a, vector<int>& b) {
+            return a[1] < b[1];
+        });
+        int ans = 0;
+        long long last = -(1LL << 60);
+        for (auto& p : points) {
+            int a = p[0], b = p[1];
+            if (a > last) {
+                ++ans;
+                last = b;
+            }
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+func findMinArrowShots(points [][]int) (ans int) {
+	sort.Slice(points, func(i, j int) bool { return points[i][1] < points[j][1] })
+	last := -(1 << 60)
+	for _, p := range points {
+		a, b := p[0], p[1]
+		if a > last {
+			ans++
+			last = b
+		}
+	}
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function findMinArrowShots(points: number[][]): number {
+    points.sort((a, b) => a[1] - b[1]);
+    let ans = 0;
+    let last = -Infinity;
+    for (const [a, b] of points) {
+        if (last < a) {
+            ans++;
+            last = b;
+        }
+    }
+    return ans;
+}
+```
+
+#### C#
+
+```cs
+public class Solution {
+    public int FindMinArrowShots(int[][] points) {
+        Array.Sort(points, (a, b) => a[1] < b[1] ? -1 : a[1] > b[1] ? 1 : 0);
+        int ans = 0;
+        long last = long.MinValue;
+        foreach (var point in points) {
+            if (point[0] > last) {
+                ++ans;
+                last = point[1];
+            }
+        }
+        return ans;
+    }
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->