10. Minimum Number of Arrows to Burst Balloons.cpp

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

bool comp(vector<int> &a, vector<int> &b)
{
  if (a[1] == b[1])
    return a[0] < b[0];
  return a[1] < b[1];
}

class Solution
{
public:
  int findMinArrowShots(vector<vector<int>> &points)
  {
    if (points.size() == 0)
      return 0;
    sort(points.begin(), points.end(), comp);
    int ans = 1;
    int end = points[0][1];
    for (int i = 1; i < points.size(); ++i)
    {
      if (points[i][0] > end)
      {
        ans++;
        end = points[i][1];
      }
    }
    return ans;
  }
};