leetcode 452 - Minimum Number of Arrows to Burst Balloons

https://leetcode.com/problems/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 <= 10^4
points[i].length == 2
-2^31 <= xstart < xend <= 2^31 - 1

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class Solution {
public:
int findMinArrowShots(vector<vector<int>>& points) {
int size = points.size();
if (size == 0) return 0;

auto mycomp = [](const vector<int> &p1, const vector<int> &p2) {
return p1[0] < p2[0];
};

sort(points.begin(), points.end(), mycomp);

int shots = 1;
vector<int> cur_points = points[0];

for (int i = 1; i < size; i++) {
if (points[i][0] > cur_points[1]) {
shots++;
cur_points = points[i];
} else {
cur_points[0] = points[i][0];
cur_points[1] = min(cur_points[1], points[i][1]);
}
}

return shots;
}
};