feat: add minimum-number-of-arrows-to-burst-balloons

Change-Id: I938311de832bdb93b3ffd95d1add3ebe57c8efd5

README.md

@@ -233,6 +233,7 @@ LeetCode is a very good website to sharpen your programming/problem-solving skil
 |445|[Add Two Numbers II](https://leetcode.com/problems/add-two-numbers-ii/)|[Python](./algorithms/add-two-numbers-ii/)|![Medium](https://img.shields.io/badge/-Medium-orange)|
 |448|[Find All Numbers Disappeared in an Array](https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/)|[Java](./algorithms/find-all-numbers-disappeared-in-an-array/)|![Easy](https://img.shields.io/badge/-Easy-green)|
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+|452|[Minimum Number of Arrows to Burst Balloons](https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/)|[Python](./algorithms/minimum-number-of-arrows-to-burst-balloons/)|![Medium](https://img.shields.io/badge/-Medium-orange)|
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algorithms/minimum-number-of-arrows-to-burst-balloons/Solution.py

@@ -0,0 +1,53 @@
+"""
+There are some spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter, and hence the x-coordinates of start and end of the diameter suffice. The start is always smaller than the end.
+
+An arrow can be shot up exactly vertically from different points along the x-axis. A balloon with xstart and xend bursts by an arrow shot at x if xstart ≤ x ≤ xend. There is no limit to the number of arrows that can be shot. An arrow once shot keeps traveling up infinitely.
+
+Given an array points where points[i] = [xstart, xend], 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: One way is to shoot one arrow for example at x = 6 (bursting the balloons [2,8] and [1,6]) and another arrow at x = 11 (bursting the other two balloons).
+Example 2:
+
+Input: points = [[1,2],[3,4],[5,6],[7,8]]
+Output: 4
+Example 3:
+
+Input: points = [[1,2],[2,3],[3,4],[4,5]]
+Output: 2
+Example 4:
+
+Input: points = [[1,2]]
+Output: 1
+Example 5:
+
+Input: points = [[2,3],[2,3]]
+Output: 1
+ 
+
+Constraints:
+
+0 <= points.length <= 104
+points.length == 2
+-231 <= xstart < xend <= 231 - 1
+"""
+
+
+class Solution:
+    def findMinArrowShots(self, points: List[List[int]]) -> int:
+        points.sort(key=lambda x:(x[1],x[0]))
+        prev, ret = None, 0
+        for p in points:
+            if prev is None:
+                prev = p[1]
+                ret += 1
+                continue
+            if prev < p[0]:
+                prev = p[1]
+                ret += 1
+        return ret