10. 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

Solution: (Greedy )

Approach: Finding no of overlapping intervals and counting no of arrows needed

class Solution
{
public:
    static bool comparator(vector<int> &a, vector<int> &b)
    {
        return a[1] < b[1];
    }

    int findMinArrowShots(vector<vector<int>> &points)
    {

        sort(points.begin(), points.end(), comparator);
        int n = points.size();
        if (n == 0)
        {
            return 0;
        }
        int i = 0, j = 1;
        int count = 1;
        while (j < n)
        {
            if (points[j][0] <= points[i][1])
            {
                j++;
            }
            else
            {
                count++;
                i = j;
                j++;
            }
        }

        return count;
    }
};

Time Complexity: O(n log n)