Maximum_Number_Of_Arrows_To_Burst_Balloons.py

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

# O(n) Time and O(1) Space
class Solution:
    def findMinArrowShots(self, points: List[List[int]]) -> int:
        
        if len(points) == 0:
            return 0
        
        points.sort(key=lambda x: x[1])
        
        # print(points)
        
        result = 1
        end = points[0][1]
        
        for i in range(1, len(points)):
            
            if points[i][0] > end:
                result += 1
                end = points[i][1]
        
        return result