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

452. Minimum Number of Arrows to Burst Balloons
struct Solution;

impl Solution {
fn find_min_arrow_shots(mut points: Vec<Vec<i32>>) -> i32 {
let n = points.len();
if n == 0 {
return 0;
}
points.sort_by_key(|p| p[1]);
let mut end = points[0][1];
let mut res = 1;
for i in 1..n {
if points[i][0] <= end {
continue;
}
end = points[i][1];
res += 1;
}
res
}
}

#[test]
fn test() {
let points = vec_vec_i32![[10, 16], [2, 8], [1, 6], [7, 12]];
let res = 2;
assert_eq!(Solution::find_min_arrow_shots(points), res);
}