diff --git a/Solution/452. Minimum Number of Arrows to Burst Balloons/452. Minimum Number of Arrows to Burst Balloons.py b/Solution/452. Minimum Number of Arrows to Burst Balloons/452. Minimum Number of Arrows to Burst Balloons.py
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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
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diff --git a/Solution/452. Minimum Number of Arrows to Burst Balloons/readme.md b/Solution/452. Minimum Number of Arrows to Burst Balloons/readme.md
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+---
+comments: true
+difficulty: Medium
+edit_url: Antim
+tags:
+ - Greedy
+ - Array
+ - Sorting
+---
+
+
+
+# [452. Minimum Number of Arrows to Burst Balloons](https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons)
+
+## Description
+
+
+
+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] = [xstart, xend] denotes a balloon whose horizontal diameter stretches between xstart and 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 xstart and xend is burst by an arrow shot at x if 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 <= points.length <= 105points[i].length == 2-231 <= xstart < xend <= 231 - 1
+
+
+
+## Solutions
+
+
+
+### Solution 1
+
+
+
+#### 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>& points) {
+ sort(points.begin(), points.end(), [](vector& a, vector& 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;
+ }
+};
+```
+
+
+
+
+
+
+
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