LC452. Minimum Number of Arrows to Burst Balloons

class Solution:
    def findMinArrowShots(self, A: List[List[int]]) -> int:
        """
        time O(NlgN) space O(1)
        0. check edge case if points is empty 
        1. sort by end point
        2. cache 1st end, start looping with 2nd 
        3. current start ptr > end  s cnt +1
        note here if end with start overlaps, one arrow will be fine 
        """
        """
        convert problem into find the most overlaped points 
        => need to shoot right most as possible =>sort  by end 
        
        """
        if not A:
            return 0
        A.sort(key=lambda x:x[1])
        end =  A[0][1]
        cnt = 1
        for s,e in A:
            if s <= end:  #can be skipped, just to make clear
                continue
            else:  
                # if s > end 
                cnt += 1
                end = e
        return cnt