Leetcode 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
[Think]
- it’s like sliding window problem, and we can use greedy algorithm to solve
- sort the array by the ending point, and loop the array
- if next ballon starting point before previous ending point, we can shoot them together.
- to avoid overflow, use Integer.compare(a[1], b[1]), instead of (a[1]-b[1])