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);
let mut end = points;
let mut res = 1;
for i in 1..n {
if points[i] <= end {
continue;
}
end = points[i];
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);
}
``````