Day-10-Minimum Number of Arrows to Burst Balloons.cpp

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.length == 2
-231 <= xstart < xend <= 231 - 1








class Solution {
public:

    static bool comp(vector<int> a, vector<int> b){
        return a[0]<b[0];
    }

    int findMinArrowShots(vector<vector<int>>& points) {
        int n=points.size();
        if(n==0) return 0;
        int arrow=1;
        sort(points.begin(), points.end(), comp);
        int end=points[0][1];
        for(int i=1;i<n;i++){
            if(points[i][0]>end){
                arrow++;
                end=points[i][1];
            } else end=min(end, points[i][1]);
        }
        return arrow;
    }
};