452. Minimum Number of Arrows to Burst Balloons
Medium
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.
Example 1:
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].
Example 2:
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.
Example 3:
Input: points = [[1,2],[2,3],[3,4],[4,5]]
Output: 2
Explanation: The balloons can be burst by 2 arrows:
- Shoot an arrow at x = 2, bursting the balloons [1,2] and [2,3].
- Shoot an arrow at x = 4, bursting the balloons [3,4] and [4,5].
Constraints:
1 <= points.length <= 105
points[i].length == 2
-231 <= xstart < xend <= 231 - 1
class Solution:
def findMinArrowShots(self, points: List[List[int]]) -> int:
points.sort(key = lambda x: x[1])
# print(points)
common = []
for point in points:
if len(common) == 0:
common.append(point)
continue
last = common[-1]
if last[1] >= point[0]:
temp = [max(point[0], last[0]), min(point[1], last[1])]
if temp[0] <= last[0] or temp[1] <= last[1]:
common[-1] = temp
else:
common.append(temp)
else:
common.append(point)
# print('->', common)
return len(common)
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