# Leetcode 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] = [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.

```
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].
```

```
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.
```

- We use greedy to solve this question. We sort the given input list by end time. We set a left boundary and Now we only need to compare the current start time and previous end time. If the current start time is greater than previous end time. Then we now we need one more arrow and we have to update the previous end time to current end time. Otherwise the current and previous ballons can be bursted by one arrow.

```
class Solution:
def findMinArrowShots(self, points: List[List[int]]) -> int:
if not points: return 0
points.sort(key=lambda x: x[1])
first_end = points[0][1]
arrow = 1
for i in range(1,len(points)):
start, end = points[i]
if start > first_end:
arrow+=1
first_end = end
return arrow
```