Amazon
2020-10-01
452. Minimum Number of Arrows to Burst Balloons
Question:
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 <= 104points.length == 2-231 <= xstart < xend <= 231 - 1
Solution:
Sort based on the destination, check if next start location is falling in the same range.
class Solution {
public int findMinArrowShots(int[][] points) {
if (points.length == 0) return 0;
Arrays.sort(points, new Comparator<int[]>(){
@Override
public int compare(int[] a, int[] b){
return a[1] >= b[1] ? 1 : -1;
}
});
int maxEnd = points[0][1];
int counter = 1;
for (int[] point: points) {
if (point[0] > maxEnd) {
counter++;
maxEnd = point[1];
}
}
return counter;
}
}