452. Minimum Number of Arrows to Burst Balloons

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[i].length == 2
• -231 <= xstart < xend <= 231 - 1

Analysis:

很像Non-overlapping interval这道题。把interval通过ending value sort一遍，然后找到ballon之间的common interval，如果下一个balloon不在这个interval之内，则加一个arrow，建立新的interval

Solution：

class Solution(object):
 def findMinArrowShots(self, points):
 “””
 :type points: List[List[int]]
 :rtype: int
 “””
 if not points: return 0
 points.sort(key = lambda x: x[1])
 ans = 1
 high = points[0][1]
 for i in range(1, len(points)):
   if points[i][0] > high:
   ans += 1
   high = points[i][1]
 
 return ans